Contents
1 Applications of Derivatives
1.1 Critical Values . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.1.1 Critical Values of Polynomials Functions . . . . . . . . .
1.1.2 Critical Values of Trigonometric Functions . . . . . . . .
1.1.3 Critical of Exponential Functions . . . . . . . . . . . . .
1.1.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.2 Increasing and Decreasing Functions . . . . . . . . . . . . . . .
1.2.1 First Derivative Test . . . . . . . . . . . . . . . . . . . .
1.2.2 Testing Polynomial Functions . . . . . . . . . . . . . . .
1.2.3 Testing Trigonometric and Logarithmic Function . . . .
1.2.4 Testing Exponential Functions . . . . . . . . . . . . . . .
1.2.5 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3 Extrema . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.3.1 Types of Extrema Points . . . . . . . . . . . . . . . . . .
1.3.2 Finding Extrema . . . . . . . . . . . . . . . . . . . . . .
1.3.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.4 More Extrema Points . . . . . . . . . . . . . . . . . . . . . . . .
1.4.1 Extrema on a Closed Interval . . . . . . . . . . . . . . .
1.4.2 Examples of Extrema on a closed interval . . . . . . . .
1.4.3 Extrema of Trigonometric and Exponential Functions . .
1.4.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5 Concavity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.5.1 Tests for Concavity . . . . . . . . . . . . . . . . . . . . .
1.5.2 Testing the Concavity of a Function on an Open Interval
1.6 Optimization . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1.6.1 Maximizing and Minimizing Area . . . . . . . . . . . . .
1.6.2 Maximizing Volume . . . . . . . . . . . . . . . . . . . . .
1.6.3 Other Optimization Problems . . . . . . . . . . . . . . .
1.6.4 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . .
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1
1.1
Applications of Derivatives
Critical Values
A critical value is any value of x where the derivative of the function f (x) is zero. f ′ (x) = 0
In this section, we examine how to find the critical values of a function. We will start by
finding the critical values of a polynomial function. The critical value of f(x) can indicate a
point where the graph has a peek or valley.
1.1.1
Critical Values of Polynomials Functions
In each of these examples, we find the critical value of the function by taking the derivative of
the function and set the derivative equal to zero. Once we setting the derivative of the function each to zero, we can find the critical values by solving the equation for the given variable.
Example 1
Find the critical values for the function f (x) = 3x2 − 4x + 2
Solution:
f (x) = 3x2 − 4x + 2
f ′ (x) = 6x − 4
Now, set the derivative equal to zero
f ′ (x) = 6x − 4
6x − 4 = 0
6x − 4 + 4 = 0 + 4
6x = 4
x = 46 = 23
Example 2
Find the critical values for the function f (x) = x3 − 6x
Solution:
f (x) = x3 − 6x
f ′ (x) = 3x2 − 6
2
Now, set the derivative equal to zero
f ′ (x) = 3x2 − 6
3x2 − 6 + 6 = 0 + 6
3x22 = 6
6
3x
3 = 3
2
x
√ =2√
x2 = 2√
⇒x=± 2
√
√
Therefore, the critical values of the function are − 2 and 2
Example 3
Find the critical values for the function f (x) = x4 − 4x3 + 4x2
Solution:
f (x) = x4 − 4x3 + 4x2
f ′ (x) = 4x3 − 12x2 + 8x
Now, set the derivative equal to zero
4x3 − 12x2 + 8x = 0
4x(x2 − 3x + 2) = 0
4x(x − 2)(x − 1) = 0
4x = 0 or x − 2 = 0 or x − 1 = 0
x = 0 or x = 2 or x = 1
1.1.2
Critical Values of Trigonometric Functions
Example 4
We can find the critical values of the trigonometric functions by using a similar process as the
polynomial functions where we take the derivative, set it equal to zero, and solve for x. Most
of the critical value problems involving trigonometry will be restricted to a closed interval
due to the cyclic nature of the trigonometric functions. Now, let’s look at an example of
finding the critical values of a trigonometric function.
Example 4
Find the critical values for the function f (x) = sin(x) + 4 on the interval
[0, 2π]
Solution:
3
f (x) = sin(x) + 4
f ′ (x) = cos(x)
Now, set the derivative equal to zero
cos(x) = 0
x = arccos(0)
x = π2 3π
2
1.1.3
Critical of Exponential Functions
Example 5
Find the critical values for the function f (x) = e2x − 2x
Solution:
f (x) = e2x − 2x
f ′ (x) = 2e2x − 2
Now, set the derivative equal to zero
2e2x − 2 = 0
2e2x = 2
e2x = 1
ln(e2x ) = ln(1)
2x = 0
x=0
1.1.4
Exercises
Find the critical values of the given function.
1. f (x) = x2 + 6x
2. g(x) = x3 − 15x2
3. f (x) = x2 + 4x
4. g(x) = cos(x) on the interval (0, 2π)
5. h(x) = e3x − 3x
6. f (x) = sin(4x) on the interval (0, π)
4
7. f (x) = x3 − 12x
8. g(x) = x4
9. f (x) = x4 +
16 3
x
3
+ 8x2 + 5
5
1.2
1.2.1
Increasing and Decreasing Functions
First Derivative Test
Now, let’s look at the behavior of a function in terms of its graph. We will describe a
function that is moving upward left to right as increasing, and we will describe a function
that is moving downward as decreasing. (See the illustration below:)
As it turns out that there is a connection between the derivative of a function and motion
of the function. (Increasing or Decreasing) If the function has a positive derivative on an
open interval, then the function is increasing on the interval. If the function has a negative
derivative on an open interval, then the function is decreasing on the interval.
Test for increasing and decreasing functions
Let f be a differentiable function on the interval (a, b)
1. If f (x) > 0 for all x in (a, b), then f is increasing on (a, b)
2. If f (x) < 0 for all x in (a, b), then f is decreasing on (a, b)
3. If f (x) = 0 for all x in (a, b), then f is constant on (a, b)
1.2.2
Testing Polynomial Functions
To determine whether a function is increasing or decreasing we will first take the derivative
of the function and then use the above derivative test. Now, lets do an example that uses
this derivative test to determine on what intervals the function is increasing or decreasing.
6
Example 1
Determine the intervals where the function is either increasing or decreasing. f (x) = x2 + 8x
Solution:
First, find any critical values of the function. This means that we will have to find the
derivative of the function.
f (x) = x2 + 8x
f ′ (x) = 2x − 8
Next, set the derivative of the function is equal to zero.
2x − 8 = 0
2x = 8
2x
2
=
x=4
8
2
Now, construct a table to determine whether the function is increasing or decreasing.
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, 4)
(4, ∞)
x=3
x=5
−
+
Decreasing Increasing
Note that when testing the derivative at x = 3 and x = 5, we get that the value of the sign
to be negative and positive respectively.
f ′ (3) = 2(3) − 8 = −2
f ′ (5) = 2(5) − 8 = 2
Notice that the graph of the function agrees with the results from the table above in terms
of where the function is increasing or decreasing.
7
Example 2
Determine the intervals where the function is either increasing or decreasing. f (x) = x3 − 4
Solution:
First, find any critical values of the function. This means that we will have to find the
derivative of the function.
f (x) = x3 − 4
f ′ (x) = 3x2
Next, set the derivative of the function is equal to zero and solve the equation for x
3x2 = 0
2
x
√ = 0√
x= 0
x=0
Now, construct a table to determine whether the function is increasing or decreasing.
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, 0)
(0, ∞)
x = −1
x=1
+
+
Increasing Increasing
Note that when testing the derivative at x = −1 and x = 1, we get that the value of the
sign to be positive in both cases.
f ′ (−1) = 3 · (−1)2 = 3 · 1 = 3
f ′ (−1) = 3 · (−1)2 = 3 · 1 = 3
8
Notice that the graph of the function agrees with the results from the table above in terms
of where the function is increasing on both sides of zero.
Example 3
Determine the intervals where the function is either increasing or decreasing. f (x) = x3 −9x2
Solution:
First, find any critical values of the function. This means that we will have to find the
derivative of the function.
f (x) = x3 − 9x2
f ′ (x) = 3x2 − 18x
Next, set the derivative of the function is equal to zero and solve the equation for x
3x2 − 18x = 0
3x(x − 6) = 0
3x = 0 or x − 6 = 0
x = 0 or x = 6
Now, construct a table to determine whether the function is increasing or decreasing.
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, 0)
(0, 6)
(6, ∞)
x = −1
x=5
x=7
+
−
+
Increasing Decreasing Increasing
Note that when testing the derivative at x = −1, x = 5, and x = 7, we get that the value
of the sign to be negative on the interval (0, 6), and positive on the interval (−∞, 0) and
9
(6, ∞).
f ′ (−1) = 3(−1)2 − 18 · (−1) = 3 · 1 + 18 = 3 + 18 = 21
f ′ (5) = 3(5)2 − 18 · 5 = 3 · 25 − 90 = 75 − 90 = −15
f ′ (7) = 3(7)2 − 18 · 7 = 3 · 49 − 126 = 147 − 126 = 21
1.2.3
Testing Trigonometric and Logarithmic Function
Example 4
Determine the intervals where the function is either increasing or decreasing. f (x) = cos(x)+
3 on the interval [0, 2π]
Solution:
First, find any critical values of the function. This means that we will have to find the
derivative of the function.
f (x) = cos(x) + 3
f ′ (x) = −sin(x)
Next, set the derivative of the function is equal to zero and solve the equation for x.
−sin(x) = 0
sin(x) = 0
x = arcsin(0)
x = 0, π, 2π
Now, construct a table to determine whether the function is increasing or decreasing.
Interval
Test Value
Sign of f ′ (x)
Conclusion
[0, π)
(π, 2π]
π
x= 2
x = 3π
2
+
−
Increasing Decreasing
Note that when testing the derivative at x = π2 , and x =
sign to be positive and negative respectively.
f ′ ( π2 ) = −sin( π2 ) = −1
f ′ ( 3π
) = −sin( 3π
) = −(−1) = 1
2
2
10
3π
,
2
we get that the value of the
1.2.4
Testing Exponential Functions
Example 5
Determine the intervals where the function is either increasing or decreasing. f (x) = e4x −
8x + 3
Solution:
First, find any critical values of the function. This means that we will have to find the
derivative of the function.
f (x) = e4x − 8x + 3
f ′ (x) = 4e4x − 8
Next, set the derivative of the function is equal to zero and solve the equation for x.
4e4x − 8 = 0
4e4x = 8
e4x = 2
ln(e4x ) = ln(2)
x = ln(2) ≈ .693
Now, construct a table to determine whether the function is increasing or decreasing.
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, ln(2)) (ln(2), ∞)
x=0
x=1
−
+
Decreasing Increasing
Note that when testing the derivative at x = 0, and x = 1, we get that the value of the sign
to be negative and positive respectively.
f ′ (0) = 4e4(0) − 8 = 4 · 1 − 8 = 4 − 8 = −4
f ′ (1) = 4e4(1) − 8 = 4 · e4 − 8 ≈ 210.4
Example 6
Determine the intervals where the function is either increasing or decreasing. f (x) = ex + 2
Solution:
First, find any critical values of the function. This means that we will have to find the
derivative of the function.
f (x) = ex + 2
11
f ′ (x) = ex
Next, set the derivative of the function is equal to zero and solve the resulting equation for x.
ex = 0
Notice, if we take the natural logarithm of both sides, we get the equation has no solution.
ln(ex ) = ln(0) =⇒ x = ln(0) Recall, that ln(0) is an undefined value. Therefore, the equation has no solution. =⇒ The function has no critical values.
Since the derivative of the function is always positive f ′ (x) = ex + 2 > 0 the function is
always increasing on (−∞, ∞)
1.2.5
Exercises
1. Determine if the function is increasing or decreasing at the given values for x.
f (x) = x2 + 6x at x = 0 and x = −5
2. Determine if the function is increasing or decreasing at the given values for x.
f (x) = x4 − 2x3 + 9x − 10 at x = 4 and x = −3
3. Determine if the function is increasing or decreasing at the given values for x.
f (x) = sin(3x) + 4x at x = π and x = 0
4. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. f (x) = x3
5. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. f (x) = cos(x) − 2 Test only on the interval (0, 2π)
6. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. f (x) = ex
7. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. g(x) = x2 + 5x
8. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. f (x) = x3 − 6x2
9. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. f (x) = x3 − 3x
12
1.3
1.3.1
Extrema
Types of Extrema Points
Definition of Extrema
The extrema points or extreme values of function can be described as either a maximum or
minimum. A maximum can be thought of as the highest point on a curve or ”peek”, and a
minimum can be thought of as the lowest point on the curve or ”valley”.
Let f be defined on a interval I containing c.
1. f (c) is the minimum of f on I if f (c) ≤ f (x) for all x on I.
2. f (c) is the maximum of f on I if f (c) ≥ f (x) for all x on I.
The maximum and minimum of a function on an interval are the extreme values or extrema
of the function. The maximum and minimum of a function are also the absolute maximum
and absolute minimum of a function.
An absolute maximum will be the highest point on the curve as seen below:
An absolute minimum will be the lowest point on the curve.
13
Definition of Relative Extrema
1. If there is an open interval containing c on which f (c) is a maximum, then f (c) is called
a relative maximum.
2. If there is an open interval containing c on which f (c) is a minimum, then f (c) is called
a relative minimum.
1.3.2
Finding Extrema
Example 1
Find all extrema points of the function.
f (x) = x2 − 6x
Solution:
First find the derivative of the function.
f (x) = x2 − 6x
f ′ (x) = 2x − 6
Now, set the derivative of the function equal to zero and solve for x
2x − 6 = 0
2x − 6 + 6 = 0 + 6
2x = 6
x=3
Now find the y-coordinate of critical point; f (3) = 32 − 6 · 3 = 9 − 18 = −9 Therefore, the
critical point of the function is (3, −9)
Now construct table to test where the function is increasing and decreasing. This will help
us determine the type of extrema we have.
14
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, 3)
(3, ∞)
x=2
x=4
−
+
Decreasing Increasing
The function has an absolute minimum at (3, −9)
Example 2
Find all extrema points of the function.
f (x) = x3 − 3x2
Solution:
First find the derivative of the function.
f (x) = x3 − 3x2
f ′ (x) = 3x2 − 6x
Now, set the derivative of the function equal to zero.
3x2 − 6x = 0
3x(x − 2) = 0
3x = 0 or x − 2 = 0
x = 0 or x = 2
Now find the y-coordinate of critical points; f (0) = (0)3 − 3(0)2 = 0 − 0 = 0 and f (2) =
23 − 6(2) = 8 − 12 = −4 Therefore, the critical points of the functions are (0, 0) and (2, −4)
Now construct table to test where the function is increasing and decreasing. This will help
us determine the type of extrema points we have.
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, 0)
(0, 2)
(2, ∞)
x = −1
x=1
x=3
+
−
+
Increasing Decreasing Increasing
15
Below are the values of the derivative at each test point. (x = −1, x = 1, and x = 3)
f ′ (−1) = 3(−1)2 − 6(−1) = 3 · 1 + 6 = 3 + 6 = 9
f ′ (1) = 3(1)2 − 6(1) = 3 · 1 − 6 = 3 − 6 = −3
f ′ (3) = 3(3)2 − 6(3) = 3 · 9 − 18 = 27 − 18
The function has a relative maximum at (0, 0) and a relative minimum at (2, −4)
Example 3
Find all extrema points of the function.
f (x) = x3 − 2
Solution:
First find the derivative of the function.
f (x) = x3 − 2
f ′ (x) = 3x2
Now, set the derivative of the function equal to zero and solve the equation for x.
3x2 = 0
2
x
√ =0√
x2 = 0
x=0
Now find the y-coordinate of critical point; f (3) = 03 −2 = 0−2 = −2 Therefore, the critical
point of the function is (0, −2)
Now construct table to test where the function is increasing and decreasing. This will help
us determine the type of extrema we have.
16
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, 0)
(, ∞)
x = −1
x=1
+
+
Increasing Increasing
Since the function is increasing on both sides of the critical point, the function has no extrema
points.
Example 4
Find all extrema points of the function.
g(x) = e2x + 4
Solution:
First find the derivative of the function.
g(x) = e2x + 4
g ′ (x) = 2e2x
Now, set the derivative of the function equal to zero and solve the equation for x.
2e2x = 0
e2x = 0
ln(e2x ) = ln(0)
2x = ln(0)
The equation above has no solution. Therefore, the function has no critical values or extrema.
17
Example 5
Find all extrema points of the function.
g(x) = x3 − 12x
Solution:
First find the derivative of the function.
g(x) = x3 − 12x
g ′ (x) = 3x2 − 12
Now, set the derivative of the function equal to zero and solve the equation for x.
3x2 − 12 = 0
3(x2 − 4) = 0
3(x + 2)(x − 2) = 0
x + 2 = 0 or x − 2 = 0
x = −2 or x = 2
Now find the y-coordinate of critical points; f (−2) = (−2)3 − 12(−2) = −8 + 24 = 16
and f (2) = 23 − 12(2) = 8 − 24 = −16
Therefore, the critical points of the functions are (−2, 16) and (2, −16)
Now construct table to test where the function is increasing and decreasing. This will help
us determine the type of extrema points we have.
Interval
Test Value
Sign of f ′ (x)
Conclusion
(−∞, −2)
(−2, 2)
(2, ∞)
x = −3
x=0
x=3
+
−
+
Increasing Decreasing Increasing
Here are the values of the derivatives at the each test point.
f ′ (−3) = 3 · (−3)2 − 12 = 3 · 9 − 12 = 36 − 12 = 24
f ′ (0) = 3 · (0)2 − 12 = 0 − 12 = −12
f ′ (3) = 3 · (3)2 − 12 = 3 · 9 − 12 = 36 − 12 = 24
18
19
1.3.3
Exercises
1. Find any extrema points if any of the given function. f (x) = x2 + 5x + 6
2. Find any extrema points if any of the given function. f (x) = x2 − 8
3. Find any extrema points if any of the given function. f (x) = x3 − 4x
4. Find any extrema points if any of the given function. f (x) = x3 − 6
5. Find any extrema points if of the given function. f (x) = x3 − 6x2
6. Find any extrema points if of the given function. f (x) = e5x + 4
7. Find any extrema points if of the given function. f (x) = x2 + 5x + 6
8. Find any extrema points of the given function. f (x) = x4 − 5
9. Find any extrema points of the given function. f (x) = x4 + 4x3
20
1.4
1.4.1
More Extrema Points
Extrema on a Closed Interval
On a closed interval the extrema points will occur where the derivative of the function is
zero or at an endpoint of the closed interval.
Therefore, when testing the extrema points of a function on a closed interval, you must test
where the derivative is zero and the endpoints.
1.4.2
Examples of Extrema on a closed interval
Example 1
Find the extrema points of the function f (x) = x2 + 4x on the interval [−2, 4]
The critical points will occur at the endpoints and where the derivative is zero.
First find the derivative of the function.
f ′ (x) = 2x + 4
Next, find the critical values of the function by setting the derivative of the function equal
to zero.
2x + 4 = 0
2x = −4
x = −2
Now, find the y-coordinate of the critical point: f (−2) = (−2)2 + 4 · (−2) = 4 − 8 = −4
The endpoints will give the following points:
f (−2) = (−2)2 + 4 · (−2) = −4 Same as the Critical Point
f (4) = 42 + 4 · 4 = 16 + 16 = 32
21
Critical points and endpoints
(−2, −4)
(4, 32)
Looking at the resulting graph of the function, we get the following extrema points.
Extrema Points
(−2, −4) Absolute Minimum
(4, 32) Absolute Maximum
Example 2
Find the extrema points of the function f (x) = x3 − 2 on the interval [−1, 2]
The critical points will occur at the endpoints and where the derivative is zero.
First find the derivative of the function.
f ′ (x) = 3x2
Next, find the critical values of the function by setting the derivative of the function equal
to zero.
3x2 = 0
2
x
√ =0√
x2 = 0 x = 0
Now, find the y-coordinate of the critical point: f (0) = 03 − 2 = 0 − 2 = −2
The endpoints will give the following points:
f (−1) = (−1)3 − 2 = −1 − 2 = −3
f (2) = 23 − 2 = 8 − 2 = 6
22
Critical points and endpoints
(−1, 3)
(0, −2)
(2, 6)
Extrema Points
(−1, 3) Absolute Maximum
(0, −2) Not an extrema point
(2, 6) Relative Minimum
Example 3
Find the extrema points of the function f (x) = x3 − 3x on the interval [−3, 3]
The critical points will occur at the endpoints and where the derivative is zero.
First find the derivative of the function.
f ′ (x) = 3x2 − 3
Next, find the critical values of the function by setting the derivative of the function equal
to zero.
3x2 − 3 = 0
3(x2 − 1) = 0
3(x − 1)(x + 1) x − 1 = 0 or x + 1 = 0
x = 1 or x = −1
Now, find the y-coordinate of the critical points:
f (−1) = (−1)3 − 3(−1) = −1 + 3 = 2
f (−1) = (1)3 − 3(1) = 1 − 3 = −2
23
The endpoints will give the following points:
f (−3) = (−3)3 − 3 · (−3) = −27 + 9 = −18
f (3) = 33 − 3 · 3 = 27 − 9 = 18
Critical points and endpoints
(−3, −18)
(−1, 2)
(1, −2)
(3, 18)
Extrema Points
(−3, −18) Absolute Minimum
(−1, 2) Relative Maximum
(1, −2) Relative Minimum
(3, 18) Absolute Maximum
1.4.3
Extrema of Trigonometric and Exponential Functions
Example 4
Find the extrema points of the function f (x) = sin(x) on the interval [0, 2π]
The critical points will occur at the endpoints and where the derivative is zero.
First find the derivative of the function.
24
f ′ (x) = cos(x)
Next, find the critical values of the function by setting the derivative of the function equal
to zero.
cos(x) = 0
x = cos−1 (0)
x = π2 or x =
3π
2
Now, find the y-coordinate of the critical points:
f ( π2 ) = sin( π2 ) = 1
f ( 3π
) = sin( 3π
) = −1
2
2
The endpoints will give the following points:
f (0) = sin(0) = 0
f (2π) = sin(2π) = 0
Critical points and endpoints
(0, 0)
( π2 , 1)
( 3π
, −1)
2
(2π, 0)
Extrema Points
25
(0, 0) Relative Minimum
( π2 , 1) Absolute Maximum
( 3π
, −1) Absolute Minimum
2
(2π, 0) Relative Maximum
Example 5
Find the extrema points of the function f (x) = e2x − 2 on the interval [0, 2]
The critical points will occur at the endpoints and where the derivative is zero.
First find the derivative of the function.
f ′ (x) = 2e2x
Next, find the critical values of the function by setting the derivative of the function equal
to zero.
2e2x = 0
e2x = 0
ln(e2x ) = ln(0)
2x = ln(0) ⇒ The function has no critical values.
Now, look at the endpoints of the function on the interval :
f (0) = e2(0) − 2 = e0 − 2 = −1
f (2) = e2·2 − 2 = e4 − 2 ≈ 52.5
Critical points and endpoints
(0, −1)
(2, 2e4 )
26
Extrema Points
(2, 2e4 ) Absolute Maximum
(0, −1) Absolute Minimum
1.4.4
Exercises
1. Find the extrema points of the function f (x) = x2 + 1 on the interval [−2, 2]
2. Find the extrema points of the function f (x) = x3 + 2 on the interval [−1, 2]
3. Find the extrema points of the function f (x) = x3 + 9x2 on the interval [−8, 2]
4. Find the extrema points of the function f (x) = cos(x) + 3 on the interval [0, 2π]
5. Find the extrema points of the function f (x) = sin(x) − 1 on the interval [0, 2π]
6. Find the extrema points of the function f (x) = e2x on the interval [−1, 2]
7. Find the extrema points of the function f (x) = ex + 1 on the interval [−2, 2]
27
1.5
1.5.1
Concavity
Tests for Concavity
The shape of the curve is described by its concavity. If the graph has a shape similar to a
parabola that opens upward on an interval, then the function is concave up on that interval.
If the shape of the graph is similar to a parabola that opens downward on an interval, then
the function is concave down on that interval. As it turns out, there is a connection between
the second derivative of a function and its concavity.
Concavity Test
Let f (x) be a function whose second derivative exist on I
1. If f ′′ (x) > 0 for all x on I, then f (x) is concave up for all x on I.
2. If f ′′ (x) < 0 for all x on I, then f (x) is concave dowm for all x on I.
3. If f ′′ (x) = 0 for all x on I, then the concavity test fails.
There is also a connection between the first derivative and the concavity of a function.
Concave Up: If a function f (x) is concave up, then f ′ (x) is increasing on I
Concave Down: If a function f (x) is concave up, then f ′ (x) is decreasing on I
Example 1
Determine if the function f (x) = sin(2x) is concave up or concave down at the values of
x = 0 and x = π4
Solution:
First find the second derivative of f (x)
f ′ (x) = 2cos(2x)
f ′′ (x) = −4sin(2x)
Now, test each point into the second derivative.
f ′′ (0) = −4sin(2 · 0) = −4sin(0) = 4(0) = 0, Thus, the function fails the concavity test at
x = 0.
f ′′ ( π4 ) = −4sin(2 π4 ) = −4sin( π2 ) = −4·1 = −4, Thus, the function is concave down at x = π4 .
Example 2
Determine if the function f (x) = x4 − 4x is concave up or concave down at the values of
x = 3 and x = −3
Solution:
First find the second derivative of f (x)
28
f ′ (x) = 4x3 − 4
f ′′ (x) = 12x2
Now, test each point into the second derivative.
f ′′ (3) = 12 · (3)2 = 12 · 9 = 98 > 0, Thus, the function is concave up.
f ′′ (−3) = 12 · (−3)2 = 12 · 9 = 98 > 0, Thus, the function is concave up.
1.5.2
Testing the Concavity of a Function on an Open Interval
Example 3
Determine the intervals on which the graph is concave upward or concave downward. f (x) =
x3 − 6
Solution:
First, find the second derivative of the function.
f ′ (x) = 3x2
f ′′ (x) = 6x
Next, set the derivative equal to zero and solve the equation for x.
6x = 0
⇒x=0
Now, let’s do a test for concavity.
Interval
(−∞, 0)
(0, ∞)
Test Value
x = −1
x=1
′′
Sign of f (x)
−
+
Conclusion
Concave Down Concave Up
The second derivative at x = −1 is negative and the second derivative at x = 1 is positive,
making the graph concave down on (−∞, 0) and concave up on (0, ∞)
f ′′ (−1) = 6(−1) = −6
f ′′ (1) = 6(1) = 6
29
Example 4
Determine the interval on which the graph is concave upward or concave downward. f (x) =
x3 − 3x2
Solution:
First, find the second derivative of the function.
f ′ (x) = 3x2 − 6x
f ′′ (x) = 6x − 6
Next, set the second derivative equal to zero.
6x − 6 = 0
6x = 6
6x
= 66
6
x=1
Now, let’s do a test for concavity.
Interval
(−∞, 1)
(1, ∞)
Test Value
x=0
x=2
Sign of f ′′ (x)
−
+
Conclusion
Concave Down Concave Up
Here are the values of the second derivatives.
f ′′ (0) = 6(0) − 6 = −6
f ′′ (2) = 6(2) − 6 = 12 − 6 = 6
30
Example 5
Determine the interval on which the graph is concave upward or concave downward. f (x) =
x2 + 5x + 9
Solution:
First, find the second derivative of the function.
f ′ (x) = 2x + 5
f ′′ (x) = 2
Notice that we cannot set the derivative equal to zero, but the second derivative, f ′′ (x) = 2,
is always positive. Therefore, the function is always concave up.
Determine the interval on which the graph is concave upward or concave downward. f (x) =
ex + 3
First, find the second derivative of the function.
f ′ (x) = ex
f ′′ (x) = ex
Notice that f ′′ (x) = ex > 0 Therefore, the function is always concave up.
Definition:
The inflection point of a function f (x) is any value of x where the second derivative is equal
to zero. f ′′ (x) = 0
Example 7
Find any inflection points of the function f (x) = x3 + 6x2
31
Solution:
Find the second derivative of the function and set the second derivative equal to zero.
f (x) = x3 + 6x2
f ′ (x) = 3x2 + 12x
f ′′ (x) = 6x + 12
6x + 12 = 0
6x + 12 − 12 = 0 − 12
6x = −12
x = −2
Example 8
Find any inflection points of the function f (x) = 21 x4 − 6x3 + 21x2
Solution:
Find the second derivative of the function and set the second derivative equal to zero.
f (x) = 12 x4 − 6x3 + 21x2
f ′ (x) = 4 · 21 x4−1 − 3 · 6x3−1 + 2 · 24x2−1
f ′ (x) = 2x3 − 18x2 − 48x
f ′′ (x) = 6x2 − 36x − 48
6x2 − 36x − 48 = 0
6(x2 − 6x + 8) = 0
6(x − 4)(x − 2) = 0
(x − 4)(x − 2) = 0
x − 4 = 0 or x − 2 = 0
x = 4 or x = 2
32
1.6
Optimization
In basic Calculus we can use the derivative to find both maximum and minimum values.
When the derivative gives a horizontal tangent line, this will be a candidate for either a peak
(maximum) or valley (minimum). In the first portion of this section will use the derivative
to maximize values.
1.6.1
Maximizing and Minimizing Area
Example 1
Find the length and width of a rectangle has a perimeter of 140 meters and a maximum
area.
Solution:
The perimeter will be given by the formula P = 2 · l + 2 · w and the area will be given by
the formula A = l · w. Since the perimeter is 140 meters the formula for perimeter will now
be given by 2 · l + 2 · w = 140
First, solve the perimeter equation for w;
2 · l + 2 · w = 140
2 · w = 140 − 2 · l
2·w
2
= 140−2·l
2
140−2·l
w=
2
w = 70 − l
Now, substitute the value of w into the Area which results in the following of one variable.
A = l · w = l(70 − l) = 70 · l − l2
Now, maximize the function by taking its derivative.
A′ = 70 − 2l
Next, find the critical value of the area function by setting the derivative equal to zero.
70 − 2l = 0
70 = 2l
= 35 meters
l = 70
2
Therefore, the length of the maximum rectangle is 35 meters and the width is w = 70 − l =
70 − 35 = 35 meters
33
Example 2
Find the length and width of a rectangle has a area of 40 square meters and a minimum
perimeter.
Solution:
The perimeter will be given by the formula P = 2 · l + 2 · w and the area will be given by the
formula A = l · w. Since the area is 40 square meters, the formula for perimeter will now be
given by l · w = 40
First, solve the area equation for w;
l · w = 40
l·w
l
= 40l
w = 40
l
Now, substitute the value of w into the Area which results in the following of one variable.
P = 2( 40l ) + 2 · l
P = 80l + 2l˙
P = 80 · l−1 + 2 · l
Now, minimize the perimeter by taking the derivative of P
P ′ = −80 · l−2 + 2
Next, find the critical value of the area function by setting the derivative equal to zero.
−80 · l−2 + 2 = 0
= −2
− 80
l2
2 80
(l )( l2 ) = 2(l2 )
80 = 2l2
2
l√
= 40√
l2 √
= 40
l = 40 ≈ 6.3 meters
The rectangle would have a length of 6.3 meters and a width of w =
1.6.2
40
6.3
= 6.3 meters
Maximizing Volume
Example 3
A manufacturer wants to design an open box having a square base and a surface area of 108
square inches, as shown in Figure 1 − 1. What dimensions will produce a box with maximum
volume?
Figure 1.1
34
The Volume of the open box would be:
V =l·w·h
The surface area of the box would be:
A = 4·(Area of each face) + Area of the bottom.
A = 4 · xh + x · x
A = 4 · xh + x2
Since the surface area is 108 square inches, the new surface area formula would be:
108 = 4 · xh + x2
Now, solve this equation for h.
108 = 4 · xh + x2
108 − x2 = 4 · xh
108−x2
= 4xh
4x
4x
108−x2
h = 4x
Now, express the volume in terms of x.
2
V = x2 h = x2 ( 108−x
) = 41 (108 · x − x3 )
4x
Next, Minimize the volume by taking the derivative and setting the derivative equal to zero.
V = 14 (108x − x3 ) = 27x − 14 x3
V ′ = 27 − 34 x2
27 − 34 x2 = 0
35
− 34 x2 = −27
3 2
x = 27
4
x2 = 43 · 27
2
x
√ = 36√
x2 = 36
x=6
Example 4
A rectangle package to be sent by a postal service can have a maximum combined length
and girth (perimeter of a cross section) of 108 inches. Find the dimensions of the package of
maximum volume that can be sent.
First, find the formula for the girth.
Girth = 2x + 2x + y = 4x + y
108 = 4x + y
Now, solve the equation for y.
108 = 4x + y ⇒ y = 108 − 4x
Next, find the formula for the volume in terms of y.
V
V
V
V
=l·w·h=x·y·x
= x2 y
= x2 (108 − 4x)
= 108x2 − 4x3
Next, take the derivative of V and set the derivative equal to zero.
V = 108x2 − 4x3
V ′ = 216x − 12x2
216x − 12x2 = 0
12x(18 − x) = 0
12x = 0 or 18 − x = 0
36
x = 0 or x = 18
⇒ x = 18 inches
Now, find the value of y:
y = 108 − 4x = 108 − 4 · 18 = 108 − 72 = 36 inches
Thus, the package would be 18 inches by 18 inches by 36 inches.
1.6.3
Other Optimization Problems
Example 5
Find the two positive numbers that satisfy the given requirements. The product 196 and
the sum is a minimum.
Let x be the first number and y be the second number.
If the product of the numbers x and y is 196, the formula would be:
x · y = 196
Next, solve the equation for y.
x · y = 196
xy
y
196
y
196
y= x
=
Now, we will write the sum as a function of x.
S =x+y
S = x+ 196
x
S = x + 196x−1
Next, we minimize by taking the derivative.
S ′ = 1 − 196x2
0 = 1 − 196x−2
196x−2 = 1
196
x2 = 1
2
x2 · ( 196
x2 ) = 1 · x
2
x
√ = 196
√
x2 = 196
x = 14
37
1.6.4
Exercises
1. Find the critical value if any of the function f (x) = x2 − 9
2. Find the critical value if any of the function g(x) = 2x3 − 9x2
3. Find the critical value if any of the function h(x) = ex − x + 4
4. Find the critical value if any of the function f (x) = sin(6x) on the interval (0, π2 )
5. Determine if the function is increasing or decreasing at the given values for x.
f (x) = x3 − 5x at x = 0 and x = 3
6. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. g(x) = 2x4
7. Use the derivative test to determine the intervals where the function is either decreasing
or increasing. f (x) = 23 x3 + 6x2
8. Find the extrema points of the function f (x) = x2 − 3 on the interval [−1, 2]
9. Find the extrema points of the function f (x) = x3 − 4 on the interval [−1, 3]
10. Find the extrema points of the function f (x) = sin(x) + 2 on the interval [0, 2π]
11. Find the extrema points of the function f (x) = e2x − 1 on the interval [0, 1]
12. Test the concavity of the function at the given point. f (x) = −4x3 + 3x2 at x = 2
13. Determine the intervals on which the graph is concave upward or concave downward.
f (x) = x2 + 1
14. Determine the intervals on which the graph is concave upward or concave downward.
f (x) = x3 + 2x
15. Find any inflection points of the function f (x) = x4 + 6x3 + 12x2
Find any extrema points on the entire interval of real numbers.
1. Find any extrema points if they exist of the function f (x) = x2 + 5x + 3
2. Find any extrema points it they exist of the function f (x) = 2x3 − 6x2
3. Find any extrema points it they exist of the function f (x) = x3 − 6
4. Find any extrema points it they exist of the function f (x) = −x4 + 1
Page 203 #9, 11, 19, 20
38
Solutions to Review Exercises
Part 1
1. x =
9
2
2. x = 0 and x = 3
3. x = 0
4. x =
π π 5π
, ,
12 4 12
5. f ′ (0) = −5 ⇒ Decreasing: f ′ (3) = 22 ⇒ Increasing:
6. Decreasing (−∞, 0): Increasing: (0, ∞)
7. Increasing (−∞, −6): Decreasing (−6, 0): Increasing (0, ∞)
8. Relative Maximum at (−1, −2): Absolute Minimum at (0, −3): Absolute Maximum at
(2, 1)
9. Absolute Minimum at (−1, −5): Absolute Maximum at (3, 23)
10. Relative Minimum at (0, 2): Absolute Maximum at ( π2 , 3): Absolute Minimum at ( 3π
, 1):
2
Relative Maximum at (2π, 0)
11. Absolute Minimum at (0, 0): Absolute Maximum at (1, e2 − 1)
12. Concave Down at x = 2
13. Concave Up on (−∞, ∞)
14. Concave Down on (−∞, 0): Concave Up on (0, ∞)
15. Inflection Points at x = −1 and x = −2
Part 2
1. Absolute Minimum at ( 52 , 87
)
4
2. Absolute Maximum at (0, 0): Absolute Minimum at (2, −8)
3. No Extrema
4. Absolute Maximum at (0, 1)
39