   Chapter 10.2, Problem 20E ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

#### Solutions

Chapter
Section ### Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

# In Problems 21-24, a function and its first and second derivatives are given. Use these to find relative maxima, relative minima, and points of inflection; sketch the graph of each function. f ( x ) = x 5 − 5 x 4 f ' ( x ) = 5 x 2 ( x − 4 ) f ' ' ( x ) = 20 x 2 ( x − 3 )

To determine

To calculate: The relative minimum, relative maximum and points of inflection for the provided function f(x)=x55x4 and the first derivative f(x)=5x3(x4) and second derivative f(x)=20x2(x3) also sketch its graph.

Explanation

Given Information:

To find relative maxima and minima of a function,

Step 1. Set the first derivative of the function to zero, f(x)=0, to find the critical values of the function.

Step 2. Substitute the critical values into f(x) and calculate the critical points.

Step 3. Evaluate f(x) at each critical value for which f(x)=0.

If f(x0)<0, a relative maximum occurs at x0.

If f(x0)>0, a relative minimum occurs at x0.

If f(x0)=0 or f(x0) is undefined, the second derivative test fails and then use the first derivative test.

Calculation:

Consider the provided function,

f(x)=x55x4

Now, consider the first derivative f(x)=5x3(x4).

Now, to obtain the critical values, set f(x)=0 as,

5x3(x4)=0

Thus, either x4=0 or 5x3=0.

First consider x4=0.

Add 4 on both sides as below,

x4+4=0+4x=4

And,

5x3=0x3=0x=0

Thus, the critical values of the function are at x=0 and x=4.

Now, substitute 0 for x in the function f(x)=x55x4,

f(x)=(0)55(0)4=0

Substitute 4 for x in the function f(x)=x55x4,

f(x)=(4)55(4)4=10245(256)=256

Thus, the critical points are (0,0) and (4,256).

Now, consider the second derivative f(x)=20x2(x3).

To obtain point of inflection, set f(x)=0 as,

20x2(x3)=0

Thus, either x3=0 or 20x2=0.

First consider x3=0

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