   Chapter 10.1, Problem 26E ### 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

# For each function in Problems 25-30, find the relative maxima, relative minima, and horizontal points of inflection; then sketch the graph. Check your graph with a graphing utility. y = 1 4 x 4 − 2 3 x 3 + 1 2 x 2 − 2

To determine

To calculate: The relative maxima, relative minima, or horizontal points of inflection by observing the graph y=14x423x3+12x22.

Explanation

Given Information:

The provided function is y=14x423x3+12x22.

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=14x423x3+12x22,

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.

Take out the first derivative of the equation by the power rule,

y=x32x2+x=x(x22x+1)=x(x1)2

Put the value of y=0,

y=x(x1)2=0

Evaluate the values of x from the equation:

x(x1)2=0x=0,1

Hence, the value(s) of x are x=0,1.

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

Put x=0 in the equation y=14x423x3+12x22,

y=14(0)423(0)3+12(0)22=00+02=2

Hence, (0,2) is a critical point.

Put x=1 in the equation y=14x423x3+12x22,

y=14(1)423(1)3+12(1)22=1423+122=2312

Hence, (1,2312) is a critical point.

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

Left of 0 is x=1, On the right is x=12

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