“I attempted mathematics,
but it was repugnant to me,
chiefly from my not being
able to see any meaning in
the early steps in algebra.
This impatience was very
foolish, and in after years I
have deeply regretted that
I did not proceed far
enough at least to
understand something of
the great leading principles
of mathematics; for men
thus endowed seem to
have an extra sense.”
Charles Robert Darwin
(12 February 1809
- 19 April 1882)
MATH 1131Q - Calculus 1.
Álvaro Lozano-Robledo
Department of Mathematics
University of Connecticut
Day 18
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
2 / 30
Clicker Question... (frequency DD)
I do not go to your (Álvaro’s) office hours because...
(A) I don’t know when and/or where they are!
(B) I find the whole concept of “office hours” by a
professor intimidating.
(C) I don’t really have questions.
(D) I ask for help elsewhere (my TA, Q Center, etc).
(E) I can’t make it to your office at those times.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
3 / 30
Applications of Derivatives
Maximum and Minimum Values of a Function
Definition
Let c be a number in the domain D of a function f . Then f (c) is the
1
absolute maximum value of f on D if f (c) ≥ f (x) for all x in D.
2
absolute minimum value of f on D if f (c) ≤ f (x) for all x in D.
3
local maximum value of f if f (c) ≥ f (x) when x is near c.
4
local minimum value of f if f (c) ≤ f (x) when x is near c.
Maximum and Minimum Values of a Function
Definition
Let c be a number in the domain D of a function f . Then f (c) is the
1
absolute maximum value of f on D if f (c) ≥ f (x) for all x in D.
2
absolute minimum value of f on D if f (c) ≤ f (x) for all x in D.
3
local maximum value of f if f (c) ≥ f (x) when x is near c.
4
local minimum value of f if f (c) ≤ f (x) when x is near c.
Theorem (Extreme Value Theorem)
If f is continuous on a closed interval [a, b], then f attains an absolute
maximum value f (c) and an absolute minimum value f (d) at some
numbers c and d in [a, b].
Maximum and Minimum Values of a Function
Definition
Let c be a number in the domain D of a function f . Then f (c) is the
1
absolute maximum value of f on D if f (c) ≥ f (x) for all x in D.
2
absolute minimum value of f on D if f (c) ≤ f (x) for all x in D.
3
local maximum value of f if f (c) ≥ f (x) when x is near c.
4
local minimum value of f if f (c) ≤ f (x) when x is near c.
Theorem (Extreme Value Theorem)
If f is continuous on a closed interval [a, b], then f attains an absolute
maximum value f (c) and an absolute minimum value f (d) at some
numbers c and d in [a, b].
Theorem (Fermat’s Theorem)
If f has a local maximum or minimum at c, and if f 0 (c) exists, then
f 0 (c) = 0.
Maximum and Minimum Values of a Function
Theorem (Fermat’s Theorem)
If f (x) has a local maximum or minimum at c, and if f 0 (c) exists, then
f 0 (c) = 0.
Definition
A critical number of a function f (x) is a number c in the domain of f
such that (1) f 0 (c) = 0, or (2) f 0 (c) does not exist.
Example
Find the critical points of f (x) = x 3 − 3x + 1.
Maximum and Minimum Values of a Function
Theorem (Fermat’s Theorem)
If f (x) has a local maximum or minimum at c, and if f 0 (c) exists, then
f 0 (c) = 0.
Definition
A critical number of a function f (x) is a number c in the domain of f
such that (1) f 0 (c) = 0, or (2) f 0 (c) does not exist.
Example
Find the critical points of f (x) = sin x.
Maximum and Minimum Values of a Function
Example
Find the critical points of f (x) = (4 − x)x 3/ 5 .
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
8 / 30
Maximum and Minimum Values of a Function
How to find the absolute maximum and minimum values of a
function f (x) on a closed interval [a, b]:
Maximum and Minimum Values of a Function
How to find the absolute maximum and minimum values of a
function f (x) on a closed interval [a, b]:
1
Find the critical values of f in [a, b]. Calculate the values of f at all
critical values.
Maximum and Minimum Values of a Function
How to find the absolute maximum and minimum values of a
function f (x) on a closed interval [a, b]:
1
Find the critical values of f in [a, b]. Calculate the values of f at all
critical values.
2
Find the values of f at endpoints of the interval and at any point
where f is defined but not continuous.
Maximum and Minimum Values of a Function
How to find the absolute maximum and minimum values of a
function f (x) on a closed interval [a, b]:
1
Find the critical values of f in [a, b]. Calculate the values of f at all
critical values.
2
Find the values of f at endpoints of the interval and at any point
where f is defined but not continuous.
3
The largest of the values from (1) and (2) is the absolute maximum
value; the smallest of these values is the absolute minimum value.
Maximum and Minimum Values of a Function
Example
Find the absolute max and min values of f (x) = x 3 − 3x 2 + 1 in the
interval [− 12 , 4].
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
10 / 30
‚
Blue: E(t) = a/ (b
+ ce −dt ),
Red: E(t) = a +
R0
(1 + d)t
Œt
, Source
Example
Find the absolute maximum value of E(t) = a +
where a = 94, R0 = 1.077, and d = 0.0001432.
R0
(1+d)t
t
in [0, 400],
Example
Find the absolute maximum value of E(t) = a +
where a = 94, R0 = 1.077, and d = 0.0001432.
R0
(1+d)t
t
in [0, 400],
Example
Find the absolute maximum value of E(t) = a +
where a = 94, R0 = 1.077, and d = 0.0001432.
R0
(1+d)t
t
in [0, 400],
Solution: There is a unique critical point in [0, 400] given at
t=
Ln(R0 )
2Ln(1 + d)
= 259.024822948 . . .
The values of the function
E(0) = 95
E(400) ≈ 957.850
E(259.024) ≈ 14964.768.
So the absolute maximum occurs at t = 259 and the value is ≈ 14964.
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Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
14 / 30
The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be a function that satisfies the following hypotheses:
1
f is continuous on the closed interval [a, b], and
2
f is differentiable on the open interval (a, b).
Then, there is a number c in (a, b) such that
f 0 (c) =
f (b) − f (a)
b−a
, or, equivalently, f (b) − f (a) = f 0 (c)(b − a).
The Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be a function that is continuous on the closed interval [a, b], and
differentiable on the open interval (a, b). Then, there is a number c in
(a, b) such that
f 0 (c) =
f (b) − f (a)
b−a
, or, equivalently, f (b) − f (a) = f 0 (c)(b − a).
Example
Show that the Mean Value Theorem is true for f (x) = x 2 and the
interval [0, 3].
Example
A driver goes through a toll booth at noon, drives 180 miles on the
same highway, and exits through another toll 2 hours later. If the speed
limit on the highway is 65 miles per hour...
1
Was the driver speeding at some point in the trip?
2
The speed tickets are $100 if the speed at any point in time was
between 65 and 75 mi/h, and $100 + x if the speed was 75 + x
mi/h at any point. What the largest fine you can charge on the
driver?
Example
A driver goes through a toll booth at noon, drives 180 miles on the
same highway, and exits through another toll 2 hours later. If the speed
limit on the highway is 65 miles per hour...
1
Was the driver speeding at some point in the trip?
2
The speed tickets are $100 if the speed at any point in time was
between 65 and 75 mi/h, and $100 + x if the speed was 75 + x
mi/h at any point. What the largest fine you can charge on the
driver?
Important Consequences of the Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be a function that is continuous on the closed interval [a, b], and
differentiable on the open interval (a, b). Then, there is a number c in
(a, b) such that f 0 (c) = (f (b) − f (a))/ (b − a) or, equivalently,
f (b) − f (a) = f 0 (c)(b − a).
Important Consequences of the Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be a function that is continuous on the closed interval [a, b], and
differentiable on the open interval (a, b). Then, there is a number c in
(a, b) such that f 0 (c) = (f (b) − f (a))/ (b − a) or, equivalently,
f (b) − f (a) = f 0 (c)(b − a).
Theorem
If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b).
Proof.
Important Consequences of the Mean Value Theorem
Theorem (The Mean Value Theorem)
Let f be a function that is continuous on the closed interval [a, b], and
differentiable on the open interval (a, b). Then, there is a number c in
(a, b) such that f 0 (c) = (f (b) − f (a))/ (b − a) or, equivalently,
f (b) − f (a) = f 0 (c)(b − a).
Theorem
If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b).
Proof.
Let x1 and x2 be numbers in (a, b)), with a < x1 < x2 < b. Then, f is
continuous on [x1 , x2 ] and differentiable on (x1 , x2 ), because f 0 exists (and
equals 0). Hence, by the MVT there is some c in (x1 , x2 ) such that
f (x2 ) − f (x1 ) = f 0 (c)(x2 − x1 ) = 0 · (x2 − x1 ) = 0.
Therefore, f (x2 ) − f (x1 ) = 0, and so f (x2 ) = f (x1 ). It follows that f has the
same value for any two x1 and x2 on (a, b), and so it is constant.
Important Consequences of the Mean Value Theorem
Theorem
If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b).
Important Consequences of the Mean Value Theorem
Theorem
If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b).
Corollary
If f 0 (x) = g 0 (x) for all x in the interval (a, b), then f − g is constant on
(a, b); that is, f (x) = g(x) + C, where C is a constant.
Proof.
Important Consequences of the Mean Value Theorem
Theorem
If f 0 (x) = 0 for all x in an interval (a, b), then f is constant on (a, b).
Corollary
If f 0 (x) = g 0 (x) for all x in the interval (a, b), then f − g is constant on
(a, b); that is, f (x) = g(x) + C, where C is a constant.
Proof.
Consider the function F (x) = f (x) − g(x). Then,
F 0 (x) = f 0 (x) − g 0 (x) = 0 in the interval (a, b). Thus, by the theorem,
F (x) = C is constant. Hence, f (x) − g(x) = C or, equivalently,
f (x) = g(x) + C.
Sketching Graphs of
Functions
Definition
We say that a function f (x) is ...
1
Increasing in the interval [a, b], if f (c) ≤ f (d) for any c < d.
2
Decreasing in the interval [a, b], if f (c) ≥ f (d) for any c < d.
3
Concave upward in the interval [a, b], if the graph of f lies above
all of its tangent lines for points on the interval.
4
Concave downward in the interval [a, b], if the graph of f lies
below all of its tangent lines for points on the interval.
1
−3
−2
−1
0
−1
1
2
3
Using the First Derivative
Theorem (Increasing/Decreasing Test)
Let f be a differentiable function on an interval (a, b).
1
If f 0 (x) > 0 on (a, b), then f is increasing on (a, b).
2
If f 0 (x) < 0 on (a, b), then f is decreasing on (a, b).
Theorem (The First Derivative Test for Critical Points)
Suppose that c is a critical number of a continuous function f . Then:
1
2
3
If f 0 changes from positive to negative at c, then f has a local
maximum at c.
If f 0 changes from negative to positive at c, then f has a local
minimum at c.
If f 0 does not change sign at c, then f has no local maximum or
minimum at c.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
23 / 30
Using the First Derivative
Example
Sketch the graph of the function y = 3x 5 − 5x 3 + 1.
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
24 / 30
Using the First Derivative
Example
Sketch the graph of the function y = 3x 5 − 5x 3 + 1.
4
3
2
1
−1
0
1
−1
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
24 / 30
Using the First Derivative
Example
Sketch the graph of the function y = 3x 5 − 5x 3 + 1.
4
3
2
1
−1
0
1
−1
Álvaro Lozano-Robledo (UConn)
MATH 1131Q - Calculus 1
25 / 30