   # In Problems 31-36, both a function and its derivative are given. Use them to find critical values, critical points, intervals on which the function is increasing and decreasing, relative maxima, relative minima, and horizontal points of inflection; sketch the graph of each function. f ( x ) = x − 3 x 2 / 3 f ' ( x ) = x 1 / 3 − 2 x 1 / 3 ### Mathematical Applications for the ...

12th Edition
Ronald J. Harshbarger + 1 other
Publisher: Cengage Learning
ISBN: 9781337625340

#### Solutions

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

12th Edition
Ronald J. Harshbarger + 1 other
Publisher: Cengage Learning
ISBN: 9781337625340
Chapter 10.1, Problem 36E
Textbook Problem
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## In Problems 31-36, both a function and its derivative are given. Use them to find critical values, critical points, intervals on which the function is increasing and decreasing, relative maxima, relative minima, and horizontal points of inflection; sketch the graph of each function. f ( x ) = x − 3 x 2 / 3    f ' ( x ) = x 1 / 3 − 2 x 1 / 3

To determine

To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of the function y=x3x23 and the derivative dydx=x132x13 and also sketch the graph.

### Explanation of Solution

Given Information:

The provided function is y=x3x23 and the derivative is dydx=x132x13.

Formula Used:

To obtain the relative maxima and minima of a function,

1. Find the first derivative of the function.

2. Set the derivative equal to 0 to obtain the critical points.

3. Evaluate f(x) at values of x to the left and one to the right of each critical point to develop a sign diagram.

(a) If f(x)>0 to the left and f(x)<0 to the right of the critical value, the critical point is a relative maximum point.

(b) If f(x)<0 to the left and f(x)>0 to the right of the critical value, the critical point is a relative minimum point.

Calculation:

Consider the provided function y=x3x23,

The critical values are the only values at which the graph can have turning points, the derivative cannot change sign anywhere except at the critical value.

Hence, there will no change in the values of critical values as in the derivative graph.

Consider the provided derivative,

dydx=x132x13

Put the value of y=0,

y=dydx=x132x13=0

Evaluate the values of x from the equation:

x132x13=0x=0,8

Hence, the value(s) of x are x=0,8.

Evaluate the values of the original functions with the critical values:

Put x=0 in the equation y=x3x23,

y=03(0)23=0

Hence, (0,0) is a critical point.

Put x=8 in the equation y=x3x23,

y=83(8)23=83(4)=4

Hence, (8,4) is a critical point.

Test the derivative at the left and right sides of all the critical points

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