   Chapter 10.1, Problem 7E ### 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 7-10, (a) find the critical values of the function and (b) make a sign diagram and determine the relative maxima and minima. y = 2 x 3 − 12 x 2 + 6

(a)

To determine

To calculate: The critical values of the function y=2x312x2+6.

Explanation

Given Information:

The provided equation is y=2x312x2+6.

Formula Used:

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.

This, in an interval between two critical values, the sign of the derivative at any value in the interval will be the sign of the derivative at all values in the interval.

As per the First Derivative Test,

The first derivative of the function is evaluated. The first derivative is made equal to zero in order to get the critical points.

The values of the critical values are kept inside the original function which gives the critical points. The intervals of the values of x are then evaluated for the relative maximum and minimum.

Calculation:

Consider the provided equation y=2x312x2+6,

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

(b)

To determine

To calculate: The sign diagram of the function y=2x312x2+6 and relative maxima and minima.

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