   Chapter 10.2, Problem 21E ### 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. y = x 1 / 3 ( x − 4 ) y ' = 4 ( x − 1 ) 3 x 2 / 3 y ' ' = 4 ( x + 2 ) 9 x 5 / 3

To determine

To calculate: The relative minimum, relative maximum and points of inflection for the provided equation y=x13(x4) and the first derivative y'=4(x1)3x23 and second derivative y"=4(x+2)9x53 also sketch its graph.

Explanation

Given Information:

The provided equation is y=x13(x4) and the first derivative is y'=4(x1)3x23 and second derivative is y"=4(x+2)9x53.

Formula used:

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 equation,

y=x13(x4)

Now, consider the first derivative y'=4(x1)3x23.

Now, to obtain the critical values, set y'=0 as,

4(x1)3x23=0

Consider, if the denominator is 0, the function is not defined.

3x23=0x=0

Thus, 0 is a critical point.

Consider either 4(x1)=0.

First consider 4(x1)=0.

4(x1)=0x1=0

x1+1=0+1x=1

Thus, the critical values of the equation are at x=0 and x=1.

Now, substitute 0 for x in the equation y=x13(x4),

y=x13(x4)=013(04)=0

Substitute 1 for x in the equation y=x13(x4),

y=x13(x4)=113(14)=3

Thus, the critical points are (0,0) and (1,3).

Now, consider the second derivative y"=4(x+2)9x53

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