Definition
Let f be a function and S be a set of numbers.
Definition
Let f be a function and S be a set of numbers. We say f is
increasing on S iff f (x1 ) < f (x2 ) whenever x1 < x2 and x1 , x2 are
in S.
Definition
Let f be a function and S be a set of numbers. We say f is
increasing on S iff f (x1 ) < f (x2 ) whenever x1 < x2 and x1 , x2 are
in S. We say f is decreasing on S iff f (x1 ) > f (x2 ) whenever
x1 < x2 and x1 , x2 are in S.
Definition
Let f be a function and S be a set of numbers. We say f is
increasing on S iff f (x1 ) < f (x2 ) whenever x1 < x2 and x1 , x2 are
in S. We say f is decreasing on S iff f (x1 ) > f (x2 ) whenever
x1 < x2 and x1 , x2 are in S.
Definition
We define the interior I o of an interval I as follows, where a < b.
Definition
Let f be a function and S be a set of numbers. We say f is
increasing on S iff f (x1 ) < f (x2 ) whenever x1 < x2 and x1 , x2 are
in S. We say f is decreasing on S iff f (x1 ) > f (x2 ) whenever
x1 < x2 and x1 , x2 are in S.
Definition
We define the interior I o of an interval I as follows, where a < b.
[a, b]o = (a, b)
(a, b)o = (a, b)
(a, b]o = (a, b)
[a, b)o = (a, b)
(−∞, b]o = (−∞, b)
[a, ∞)o = (a, ∞)
(−∞, b)o = (−∞, b)
(a, ∞)o = (a, ∞)
Theorem
Suppose the function f is continuous on the interval I and
differentiable on its interior I o .
Theorem
Suppose the function f is continuous on the interval I and
differentiable on its interior I o .
(1) If f 0 (x) = 0 for all x in I o , then f is constant on I .
Theorem
Suppose the function f is continuous on the interval I and
differentiable on its interior I o .
(1) If f 0 (x) = 0 for all x in I o , then f is constant on I .
(2) If f 0 (x) > 0 for all x in I o , then f is increasing on I .
Theorem
Suppose the function f is continuous on the interval I and
differentiable on its interior I o .
(1) If f 0 (x) = 0 for all x in I o , then f is constant on I .
(2) If f 0 (x) > 0 for all x in I o , then f is increasing on I .
(3) If f 0 (x) < 0 for all x in I o , then f is decreasing on I .
Theorem
Suppose the function f is continuous on the interval I and
differentiable on its interior I o .
(1) If f 0 (x) = 0 for all x in I o , then f is constant on I .
(2) If f 0 (x) > 0 for all x in I o , then f is increasing on I .
(3) If f 0 (x) < 0 for all x in I o , then f is decreasing on I .
The somewhat awkward statement of this Theorem produces
stronger conclusions than the more common version. For example,
if f is continuous on [1, 2] and has a positive derivative on (1, 2),
then our Theorem says f is increasing on the closed interval [1, 2]
while the usual version would only conclude that f is increasing on
the open interval (1, 2).
Theorem
Suppose the function f is continuous on the interval I and
differentiable on its interior I o .
(1) If f 0 (x) = 0 for all x in I o , then f is constant on I .
(2) If f 0 (x) > 0 for all x in I o , then f is increasing on I .
(3) If f 0 (x) < 0 for all x in I o , then f is decreasing on I .
The somewhat awkward statement of this Theorem produces
stronger conclusions than the more common version. For example,
if f is continuous on [1, 2] and has a positive derivative on (1, 2),
then our Theorem says f is increasing on the closed interval [1, 2]
while the usual version would only conclude that f is increasing on
the open interval (1, 2).
Throughout the proof we will suppose the function f is continuous
on the interval I and differentiable on its interior I o .
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 .
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o .
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o . So
f is continuous on [x1 , x2 ] and differentiable on (x1 , x2 ).
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o . So
f is continuous on [x1 , x2 ] and differentiable on (x1 , x2 ). So the
Mean Value Theorem applies.
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o . So
f is continuous on [x1 , x2 ] and differentiable on (x1 , x2 ). So the
Mean Value Theorem applies. So there is c between x1 and x2 ,
with
f (x2 ) − f (x1 )
= f 0 (c) = 0
x2 − x1
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o . So
f is continuous on [x1 , x2 ] and differentiable on (x1 , x2 ). So the
Mean Value Theorem applies. So there is c between x1 and x2 ,
with
f (x2 ) − f (x1 )
= f 0 (c) = 0
x2 − x1
and we get the last equality since c is in I o since c is between x1
and x2 , and they are in the interval I .
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o . So
f is continuous on [x1 , x2 ] and differentiable on (x1 , x2 ). So the
Mean Value Theorem applies. So there is c between x1 and x2 ,
with
f (x2 ) − f (x1 )
= f 0 (c) = 0
x2 − x1
and we get the last equality since c is in I o since c is between x1
and x2 , and they are in the interval I . So f (x2 ) − f (x1 ) = 0 and
f (x2 ) = f (x1 ).
Proof in case f 0 (x) = 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . So the closed interval [x1 , x2 ] is
contained in I , and the open interval (x1 , x2 ) is contained in I o . So
f is continuous on [x1 , x2 ] and differentiable on (x1 , x2 ). So the
Mean Value Theorem applies. So there is c between x1 and x2 ,
with
f (x2 ) − f (x1 )
= f 0 (c) = 0
x2 − x1
and we get the last equality since c is in I o since c is between x1
and x2 , and they are in the interval I . So f (x2 ) − f (x1 ) = 0 and
f (x2 ) = f (x1 ). Since x1 and x2 were arbitrary members of I with
x1 < x2 , this means f is constant on I .
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o .
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ).
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ). So the Mean Value Theorem applies.
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ). So the Mean Value Theorem applies.
So there is c between x1 and x2 , with
f (x2 ) − f (x1 )
= f 0 (c) > 0
x2 − x1
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ). So the Mean Value Theorem applies.
So there is c between x1 and x2 , with
f (x2 ) − f (x1 )
= f 0 (c) > 0
x2 − x1
and we get the last inequality since c is in I o as above.
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ). So the Mean Value Theorem applies.
So there is c between x1 and x2 , with
f (x2 ) − f (x1 )
= f 0 (c) > 0
x2 − x1
and we get the last inequality since c is in I o as above. Since
x2 − x1 > 0, we multiply the inequality by x2 − x1 ,
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ). So the Mean Value Theorem applies.
So there is c between x1 and x2 , with
f (x2 ) − f (x1 )
= f 0 (c) > 0
x2 − x1
and we get the last inequality since c is in I o as above. Since
x2 − x1 > 0, we multiply the inequality by x2 − x1 , and we now
have f (x2 ) − f (x1 ) > 0 and f (x2 ) > f (x1 ).
Proof in case f 0 (x) > 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) < f (x2 ).)
So the closed interval [x1 , x2 ] is contained in I , and the open
interval (x1 , x2 ) is contained in I o . So f is continuous on [x1 , x2 ]
and differentiable on (x1 , x2 ). So the Mean Value Theorem applies.
So there is c between x1 and x2 , with
f (x2 ) − f (x1 )
= f 0 (c) > 0
x2 − x1
and we get the last inequality since c is in I o as above. Since
x2 − x1 > 0, we multiply the inequality by x2 − x1 , and we now
have f (x2 ) − f (x1 ) > 0 and f (x2 ) > f (x1 ). Since x1 and x2 were
arbitrary members of I with x1 < x2 , this means f is increasing on
I.
Proof in case f 0 (x) < 0 for all x in I o .
Let x1 , x2 be in I with x1 < x2 . (We aim to show f (x1 ) > f (x2 ).)
The details are similar to the previous case and are left to you.
This ends the proof of the theorem.
A Remark
If there are numbers a, b, c, with a < c < b and such that f is
increasing on (a, c] and decreasing on [c, b), then f (c) is a local
maximum value!
A Remark
If there are numbers a, b, c, with a < c < b
increasing on (a, c] and decreasing on [c, b),
maximum value!
If there are numbers a, b, c, with a < c < b
decreasing on (a, c] and increasing on [c, b),
minimum value!
and such that f is
then f (c) is a local
and such that f is
then f (c) is a local
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
If −1 < x < 1, then x − 1 < 0 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) < 0; i.e., f 0 (x) < 0 on (−1, 1).
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
If −1 < x < 1, then x − 1 < 0 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) < 0; i.e., f 0 (x) < 0 on (−1, 1).
If 1 < x, then 0 < x − 1 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
If −1 < x < 1, then x − 1 < 0 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) < 0; i.e., f 0 (x) < 0 on (−1, 1).
If 1 < x, then 0 < x − 1 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
If −1 < x < 1, then x − 1 < 0 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) < 0; i.e., f 0 (x) < 0 on (−1, 1).
If 1 < x, then 0 < x − 1 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is increasing on (−∞, −1],
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
If −1 < x < 1, then x − 1 < 0 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) < 0; i.e., f 0 (x) < 0 on (−1, 1).
If 1 < x, then 0 < x − 1 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
An Example
Problem. f (x) = x 3 − 3x + 2. Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 3x 2 − 3 = 3(x 2 − 1) = 3(x + 1)(x − 1).
So f 0 (x) = 0 iff x = −1 or 1.
If x < −1, then x − 1 < x + 1 < 0, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (−∞, −1).
If −1 < x < 1, then x − 1 < 0 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) < 0; i.e., f 0 (x) < 0 on (−1, 1).
If 1 < x, then 0 < x − 1 < x + 1, and so
f 0 (x) = 3(x + 1)(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
f is increasing on [1, +∞).
An Example, continued
We know now that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
f is increasing on [1, +∞).
An Example, continued
We know now that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
f is increasing on [1, +∞).
This latter information tells us that
f (−1) is a local maximum value, and
An Example, continued
We know now that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
f is increasing on [1, +∞).
This latter information tells us that
f (−1) is a local maximum value, and
f (1) is a local minimum value.
An Example, continued
We know now that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
f is increasing on [1, +∞).
This latter information tells us that
f (−1) is a local maximum value, and
f (1) is a local minimum value.
f (−1) = −1 + 3 + 2 = 4 and f (1) = 1 − 3 + 2 = 0.
An Example, continued
We know now that
f is increasing on (−∞, −1],
f is decreasing on [−1, 1], and
f is increasing on [1, +∞).
This latter information tells us that
f (−1) is a local maximum value, and
f (1) is a local minimum value.
f (−1) = −1 + 3 + 2 = 4 and f (1) = 1 − 3 + 2 = 0.
10
8
6
4
2
K2
K1
0
K2
K4
K6
1
x
2
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
If 0 < x < 1, then x − 1 < 0 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (0, 1).
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
If 0 < x < 1, then x − 1 < 0 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (0, 1).
If 1 < x, then 0 < x − 1 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
If 0 < x < 1, then x − 1 < 0 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (0, 1).
If 1 < x, then 0 < x − 1 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, −1],
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
If 0 < x < 1, then x − 1 < 0 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (0, 1).
If 1 < x, then 0 < x − 1 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
If 0 < x < 1, then x − 1 < 0 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (0, 1).
If 1 < x, then 0 < x − 1 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and
A Second Example
Problem. f (x) = (x 2 − 1)2 . Find where f is increasing and
decreasing. Find all local extrema. Sketch the graph.
f 0 (x) = 2(x 2 − 1)(2x) = 4x(x 2 − 1) = 4(x + 1)x(x − 1).
So f 0 (x) = 0 iff x = −1, 0, or 1.
If x < −1, then x − 1 < x < x + 1 < 0, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (−∞, −1).
If −1 < x < 0, then x − 1 < x < 0 < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (−1, 0).
If 0 < x < 1, then x − 1 < 0 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) < 0; i.e., f 0 (x) < 0 on (0, 1).
If 1 < x, then 0 < x − 1 < x < x + 1, and so
f 0 (x) = 4(x + 1)x(x − 1) > 0; i.e., f 0 (x) > 0 on (1, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
A Second Example, Continued
We know that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
A Second Example, Continued
We know that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
This tells us that
f (−1) is a local minimum value, and
A Second Example, Continued
We know that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
This tells us that
f (−1) is a local minimum value, and
f (0) is a local maximum value, and
A Second Example, Continued
We know that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
This tells us that
f (−1) is a local minimum value, and
f (0) is a local maximum value, and
f (1) is a local minimum value.
A Second Example, Continued
We know that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
This tells us that
f (−1) is a local minimum value, and
f (0) is a local maximum value, and
f (1) is a local minimum value.
f (−1) = ((−1)2 − 1)2 = 02 = 0 = f (1) and f (0) = (−1)2 = 1.
A Second Example, Continued
We know that
f is decreasing on (−∞, −1], f is increasing on [−1, 0],
f is decreasing on [0, 1], and f is increasing on [1, +∞).
This tells us that
f (−1) is a local minimum value, and
f (0) is a local maximum value, and
f (1) is a local minimum value.
f (−1) = ((−1)2 − 1)2 = 02 = 0 = f (1) and f (0) = (−1)2 = 1.
4
3
y
2
1
K2
K1
0
K1
1
x
2
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
If 0 < x and x 6= 2, then f 0 (x) = 12x(x − 2)2 > 0.
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
If 0 < x and x 6= 2, then f 0 (x) = 12x(x − 2)2 > 0.
So f 0 (x) > 0 on (0, 2), and
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
If 0 < x and x 6= 2, then f 0 (x) = 12x(x − 2)2 > 0.
So f 0 (x) > 0 on (0, 2), and f 0 (x) > 0 on (2, +∞).
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
If 0 < x and x 6= 2, then f 0 (x) = 12x(x − 2)2 > 0.
So f 0 (x) > 0 on (0, 2), and f 0 (x) > 0 on (2, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, 0],
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
If 0 < x and x 6= 2, then f 0 (x) = 12x(x − 2)2 > 0.
So f 0 (x) > 0 on (0, 2), and f 0 (x) > 0 on (2, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, 0],
f is increasing on [0, 2], and f is increasing on [2, +∞).
A Third Example
Problem. f (x) = 3x 4 − 16x 3 + 24x 2 − 10. Find where f is
increasing and decreasing. Find all local extrema. Sketch the
graph.
f 0 (x) = 12x 3 − 48x 2 + 48x = 12x(x 2 − 4x + 4) = 12x(x − 2)2 .
So f 0 (x) = 0 iff x = 0, or 2.
Note that (x − 2)2 > 0 except when x = 2.
If x < 0, then x < 0 and x 6= 2, and so f 0 (x) = 12x(x − 2)2 < 0;
i.e., f 0 (x) < 0 on (−∞, 0).
If 0 < x and x 6= 2, then f 0 (x) = 12x(x − 2)2 > 0.
So f 0 (x) > 0 on (0, 2), and f 0 (x) > 0 on (2, +∞).
Now the Theorem tells us that
f is decreasing on (−∞, 0],
f is increasing on [0, 2], and f is increasing on [2, +∞).
Since the latter two intervals overlap at 2, we have that f is
increasing on [0, +∞).
A Third Example, Continued
We know f is decreasing on (−∞, 0], and
f is increasing on [0, +∞).
A Third Example, Continued
We know f is decreasing on (−∞, 0], and
f is increasing on [0, +∞).
This tells us that
f (0) is a local and absolute minimum value.
A Third Example, Continued
We know f is decreasing on (−∞, 0], and
f is increasing on [0, +∞).
This tells us that
f (0) is a local and absolute minimum value.
f has no local maxima.
A Third Example, Continued
We know f is decreasing on (−∞, 0], and
f is increasing on [0, +∞).
This tells us that
f (0) is a local and absolute minimum value.
f has no local maxima.
f (0) = −10 and
f (2) = 3·16−16·8+24·4−10 = −5·16+6·16−10 = 16−10 = 6.
A Third Example, Continued
We know f is decreasing on (−∞, 0], and
f is increasing on [0, +∞).
This tells us that
f (0) is a local and absolute minimum value.
f has no local maxima.
f (0) = −10 and
f (2) = 3·16−16·8+24·4−10 = −5·16+6·16−10 = 16−10 = 6.
15
10
y
5
K1
0
K5
K10
1
2
x
3
4
5
Another Example and a WARNING
If f (x) = x1 , then f 0 (x) =
−1
x2
< 0 for all x 6= 0.
Another Example and a WARNING
If f (x) = x1 , then f 0 (x) =
(−∞, 0) and on (0, ∞).
−1
x2
< 0 for all x 6= 0. f is continuous on
Another Example and a WARNING
< 0 for all x 6= 0. f is continuous on
If f (x) = x1 , then f 0 (x) = −1
x2
(−∞, 0) and on (0, ∞). So by the Theorem, f is decreasing on
(−∞, 0) and on (0, ∞).
Another Example and a WARNING
< 0 for all x 6= 0. f is continuous on
If f (x) = x1 , then f 0 (x) = −1
x2
(−∞, 0) and on (0, ∞). So by the Theorem, f is decreasing on
(−∞, 0) and on (0, ∞).
Warning
6
1
x
f (x) = is NOT decreasing
on (−∞, 0) ∪ (0, ∞) because, for
example, f (−1) = −1 < 1 = f (1), but if
it were decreasing on (−∞, 0) ∪ (0, ∞)
we should instead have f (−1) > f (1).
4
y
2
K6
K4
K2
0
2
x
4
6
K2
K4
K6
Figure: The graph of 1/x
You know that
2x =
d 2
d 2
d 2
d 2
d 2 √
(x ) =
(x −1) =
(x +17) =
(x + 2) =
(x +C )
dx
dx
dx
dx
dx
where C is any constant.
You know that
2x =
d 2
d 2
d 2
d 2
d 2 √
(x ) =
(x −1) =
(x +17) =
(x + 2) =
(x +C )
dx
dx
dx
dx
dx
where C is any constant.
Is there any other sort of solution to
dy
= 2x?
dx
You know that
2x =
d 2
d 2
d 2
d 2
d 2 √
(x ) =
(x −1) =
(x +17) =
(x + 2) =
(x +C )
dx
dx
dx
dx
dx
where C is any constant.
Is there any other sort of solution to
dy
= 2x?
dx
The next Theorem says, “No, there is not.”
Theorem
Suppose the functions f and g are continuous on the interval I . If
f 0 (x) = g 0 (x) for all x in its interior I o , then there is a constant C
so that
g (x) = f (x) + C
for all x in I .
Theorem
Suppose the functions f and g are continuous on the interval I . If
f 0 (x) = g 0 (x) for all x in its interior I o , then there is a constant C
so that
g (x) = f (x) + C
for all x in I .
Proof.
Suppose f and g are as in the hypotheses. Set
h(x) = g (x) − f (x).
Theorem
Suppose the functions f and g are continuous on the interval I . If
f 0 (x) = g 0 (x) for all x in its interior I o , then there is a constant C
so that
g (x) = f (x) + C
for all x in I .
Proof.
Suppose f and g are as in the hypotheses. Set
h(x) = g (x) − f (x). Then h is continuous on I , and
h0 (x) = g 0 (x) − f 0 (x) = f 0 (x) − f 0 (x) = 0 for any x in I o .
Theorem
Suppose the functions f and g are continuous on the interval I . If
f 0 (x) = g 0 (x) for all x in its interior I o , then there is a constant C
so that
g (x) = f (x) + C
for all x in I .
Proof.
Suppose f and g are as in the hypotheses. Set
h(x) = g (x) − f (x). Then h is continuous on I , and
h0 (x) = g 0 (x) − f 0 (x) = f 0 (x) − f 0 (x) = 0 for any x in I o . Now by
(1) of the first Theorem h is constant on I ;
Theorem
Suppose the functions f and g are continuous on the interval I . If
f 0 (x) = g 0 (x) for all x in its interior I o , then there is a constant C
so that
g (x) = f (x) + C
for all x in I .
Proof.
Suppose f and g are as in the hypotheses. Set
h(x) = g (x) − f (x). Then h is continuous on I , and
h0 (x) = g 0 (x) − f 0 (x) = f 0 (x) − f 0 (x) = 0 for any x in I o . Now by
(1) of the first Theorem h is constant on I ; i.e., there is a constant
C so that g (x) − f (x) = h(x) = C for all x in I .
Theorem
Suppose the functions f and g are continuous on the interval I . If
f 0 (x) = g 0 (x) for all x in its interior I o , then there is a constant C
so that
g (x) = f (x) + C
for all x in I .
Proof.
Suppose f and g are as in the hypotheses. Set
h(x) = g (x) − f (x). Then h is continuous on I , and
h0 (x) = g 0 (x) − f 0 (x) = f 0 (x) − f 0 (x) = 0 for any x in I o . Now by
(1) of the first Theorem h is constant on I ; i.e., there is a constant
C so that g (x) − f (x) = h(x) = C for all x in I . That is,
g (x) = f (x) + C for all x in I .
Definition
F is an antiderivative of f on the interval I if F is continuous on I
and
F 0 (x) = f (x)
for all x in I o .
Definition
F is an antiderivative of f on the interval I if F is continuous on I
and
F 0 (x) = f (x)
for all x in I o .
Proposition
Any antiderivative of x n equals
x n+1
+ C,
n+1
for some constant C .
for n 6= −1
Definition
F is an antiderivative of f on the interval I if F is continuous on I
and
F 0 (x) = f (x)
for all x in I o .
Proposition
Any antiderivative of x n equals
x n+1
+ C,
n+1
for n 6= −1
for some constant C .
Proof.
d
x n+1
n+1
dx
=
1
(n + 1)x n+1−1 = x n .
n+1
Definition
F is an antiderivative of f on the interval I if F is continuous on I
and
F 0 (x) = f (x)
for all x in I o .
Proposition
Any antiderivative of x n equals
x n+1
+ C,
n+1
for n 6= −1
for some constant C .
Proof.
d
x n+1
n+1
dx
Now apply Theorem 2.
=
1
(n + 1)x n+1−1 = x n .
n+1
Examples of Antiderivatives
Example
Any antiderivative of x 4 equals
x5
5
+ C for some constant C ,
Examples of Antiderivatives
Example
5
Any antiderivative of x 4 equals x5 + C for some constant C , and
−3
any antiderivative of x −4 equals x−3 + C for some constant C .
Examples of Antiderivatives
Example
5
Any antiderivative of x 4 equals x5 + C for some constant C , and
−3
any antiderivative of x −4 equals x−3 + C for some constant C .
Example
Any antiderivative of cos x equals sin x + C for some constant C ,
Examples of Antiderivatives
Example
5
Any antiderivative of x 4 equals x5 + C for some constant C , and
−3
any antiderivative of x −4 equals x−3 + C for some constant C .
Example
Any antiderivative of cos x equals sin x + C for some constant C ,
and any antiderivative of sin x equals − cos x + C for some
d
constant C because dx
(− cos x) = − − sin x = sin x.
More on Antiderivatives
Proposition
If F and G are antiderivatives of f and g , respectively, and c is a
real number, then
More on Antiderivatives
Proposition
If F and G are antiderivatives of f and g , respectively, and c is a
real number, then
I
F + G is an antiderivative of f + g , and
More on Antiderivatives
Proposition
If F and G are antiderivatives of f and g , respectively, and c is a
real number, then
I
F + G is an antiderivative of f + g , and
I
cF is an antiderivative of cf .
More on Antiderivatives
Proposition
If F and G are antiderivatives of f and g , respectively, and c is a
real number, then
I
F + G is an antiderivative of f + g , and
I
cF is an antiderivative of cf .
Proof.
d
(F (x) + G (x)) = F 0 (x) + G 0 (x) = f (x) + g (x)
dx
More on Antiderivatives
Proposition
If F and G are antiderivatives of f and g , respectively, and c is a
real number, then
I
F + G is an antiderivative of f + g , and
I
cF is an antiderivative of cf .
Proof.
d
(F (x) + G (x)) = F 0 (x) + G 0 (x) = f (x) + g (x)
dx
and
d
(cF (x)) = cF 0 (x) = cf (x).
dx
A problem on antiderivatives
Problem: Given f 0 (x) = 5x 4 + 6x 3 − 5x 2 + 11 and f (0) = −3, find
f.
A problem on antiderivatives
Problem: Given f 0 (x) = 5x 4 + 6x 3 − 5x 2 + 11 and f (0) = −3, find
f.
Using the previous two Propositions (and Theorem 2), we see that
A problem on antiderivatives
Problem: Given f 0 (x) = 5x 4 + 6x 3 − 5x 2 + 11 and f (0) = −3, find
f.
Using the previous two Propositions (and Theorem 2), we see that
6x 4 5x 3
−
+ 11x + C
and
4
3
−3 = f (0) = 0 + 0 − 0 + 0 + C = C
and so
f (x) = x 5 +
f (x) = x 5 + 23 x 4 − 53 x 3 + 11x − 3.
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
In what follows a is acceleration, v is velocity, and p is position.
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
In what follows a is acceleration, v is velocity, and p is position.
We start with the understanding that a freely falling body near the
earth’s surface experiences a constant acceleration of 32 feet per
second per second. That is,
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
In what follows a is acceleration, v is velocity, and p is position.
We start with the understanding that a freely falling body near the
earth’s surface experiences a constant acceleration of 32 feet per
second per second. That is,
a(t) = −32,
but
v 0 (t) = a(t).
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
In what follows a is acceleration, v is velocity, and p is position.
We start with the understanding that a freely falling body near the
earth’s surface experiences a constant acceleration of 32 feet per
second per second. That is,
a(t) = −32,
but
v 0 (t) = a(t).
So
v (t) = −32t + C .
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
In what follows a is acceleration, v is velocity, and p is position.
We start with the understanding that a freely falling body near the
earth’s surface experiences a constant acceleration of 32 feet per
second per second. That is,
a(t) = −32,
but
v 0 (t) = a(t).
So
v (t) = −32t + C .
Since v (0) = 0 + C = C , we replace C by the constant v0 (which
denotes initial velocity.) That is,
Derivation of a law of physics
Now we use the above propositions to derive some laws of physics
about a freely falling body near the earth’s surface.
In what follows a is acceleration, v is velocity, and p is position.
We start with the understanding that a freely falling body near the
earth’s surface experiences a constant acceleration of 32 feet per
second per second. That is,
a(t) = −32,
but
v 0 (t) = a(t).
So
v (t) = −32t + C .
Since v (0) = 0 + C = C , we replace C by the constant v0 (which
denotes initial velocity.) That is,
v (t) = −32t + v0 ,
but
p 0 (t) = v (t).
Derivation of a law of physics, continued
So
p(t) = −16t 2 + v0 t + D.
Derivation of a law of physics, continued
So
p(t) = −16t 2 + v0 t + D.
Since p(0) = 0 + D = D, we replace D by the constant p0 (which
denotes initial position.) That is,
Derivation of a law of physics, continued
So
p(t) = −16t 2 + v0 t + D.
Since p(0) = 0 + D = D, we replace D by the constant p0 (which
denotes initial position.) That is,
p(t) = −16t 2 + v0 t + p0
where v0 is the velocity at time 0 and p0 is the position at time 0.
Example of a freely falling object
A ball is thrown upward at a speed of 16 ft./sec. from the edge of
a cliff 96 feet above the ground below. The ball is thrown so that
it lands below. How long is it till the ball hits the ground below?
Example of a freely falling object
A ball is thrown upward at a speed of 16 ft./sec. from the edge of
a cliff 96 feet above the ground below. The ball is thrown so that
it lands below. How long is it till the ball hits the ground below?
We are given that p0 = 96 and v0 = 16, and so
p(t) = −16t 2 + 16t + 96.
Example of a freely falling object
A ball is thrown upward at a speed of 16 ft./sec. from the edge of
a cliff 96 feet above the ground below. The ball is thrown so that
it lands below. How long is it till the ball hits the ground below?
We are given that p0 = 96 and v0 = 16, and so
p(t) = −16t 2 + 16t + 96. We seek the solution to 0 = p(t);
Example of a freely falling object
A ball is thrown upward at a speed of 16 ft./sec. from the edge of
a cliff 96 feet above the ground below. The ball is thrown so that
it lands below. How long is it till the ball hits the ground below?
We are given that p0 = 96 and v0 = 16, and so
p(t) = −16t 2 + 16t + 96. We seek the solution to 0 = p(t); that
is, we seek the solution to
0 = −16t 2 + 16t + 96 = −16(t 2 − t − 6) = −16(t − 3)(t + 2).
Example of a freely falling object
A ball is thrown upward at a speed of 16 ft./sec. from the edge of
a cliff 96 feet above the ground below. The ball is thrown so that
it lands below. How long is it till the ball hits the ground below?
We are given that p0 = 96 and v0 = 16, and so
p(t) = −16t 2 + 16t + 96. We seek the solution to 0 = p(t); that
is, we seek the solution to
0 = −16t 2 + 16t + 96 = −16(t 2 − t − 6) = −16(t − 3)(t + 2). So
t = 3 (or −2, but we ignore −2 since that is a time before the ball
was thrown).
Definition
Z
f (x) dx is the family of all antiderivatives of f and is called the
indefinite integral of f .
Definition
Z
f (x) dx is the family of all antiderivatives of f and is called the
indefinite integral of f .
Theorem 2 says any two antiderivatives of f (x) differ by a
constant. So we have the
Definition
Z
f (x) dx is the family of all antiderivatives of f and is called the
indefinite integral of f .
Theorem 2 says any two antiderivatives of f (x) differ by a
constant. So we have the
Proposition
Z
If F is any antiderivative of f , then
f (x) dx = F (x) + C .
Definition
Z
f (x) dx is the family of all antiderivatives of f and is called the
indefinite integral of f .
Theorem 2 says any two antiderivatives of f (x) differ by a
constant. So we have the
Proposition
Z
If F is any antiderivative of f , then
f (x) dx = F (x) + C .
The Propositions 4 and 5 above can be restated as follows.
Definition
Z
f (x) dx is the family of all antiderivatives of f and is called the
indefinite integral of f .
Theorem 2 says any two antiderivatives of f (x) differ by a
constant. So we have the
Proposition
Z
If F is any antiderivative of f , then
f (x) dx = F (x) + C .
The Propositions 4 and 5 above can be restated as follows.
Corollary
Suppose f and g have antiderivatives and c is a number.
Z
Z
Z
(1)
(f (x) + g (x))dx = f (x) dx + g (x) dx.
Z
Z
(2)
cf (x) dx = c f (x) dx.
Z
x n+1
(3) If n is constant, x n dx =
+ C , for n 6= −1.
n+1
more on indefinite integrals
Example
Z
(5x 2 + 3x + 4)dx = 5
5
x3
x2
+ 3 + 4x + C .
3
2
Z
x 2 dx + 3
Z
Z
x dx + 4
1 dx =
more on indefinite integrals
Example
Z
(5x 2 + 3x + 4)dx = 5
5
Z
x 2 dx + 3
Z
x3
x2
+ 3 + 4x + C .
3
2
Z
Also note that
cos x dx = sin x + C
Z
x dx + 4
1 dx =
more on indefinite integrals
Example
Z
(5x 2 + 3x + 4)dx = 5
5
Z
x 2 dx + 3
Z
Z
x dx + 4
x3
x2
+ 3 + 4x + C .
3
2
Z
Also note that cos x dx = sin x + C and
Z
sin x
sin x dx = − cos x + C because d dx
= cos x
1 dx =
more on indefinite integrals
Example
Z
(5x 2 + 3x + 4)dx = 5
5
Z
x 2 dx + 3
Z
Z
x dx + 4
1 dx =
x3
x2
+ 3 + 4x + C .
3
2
Z
Also note that cos x dx = sin x + C and
Z
sin x
sin x dx = − cos x + C because d dx
= cos x and because
d (− cos x)
dx
= − − sin x = sin x, etc.
Some Indefinite Integrals
Z
Z
sin x dx = − cos x + C
cos x dx = sin x + C
Z
Z
sec2 x dx = tan x + C
Z
Z
sec x tan x dx = sec x + C
Z
csc2 x dx = − cot x + C
x n dx =
x n+1
+C
n+1
for n 6= −1
csc x cot x dx = − csc x + C