   Chapter 10.1, Problem 30E ### 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 6 x 6 − x 4 + 7

To determine

To calculate: The relative maxima, minima and horizontal points of inflection for the provided function y=16x6x4+7, then sketch the graph and verify your graph with a graphing utility.

Explanation

Given Information:

The provided function is,

y=16x6x4+7

Formula used:

To obtain the relative maxima and minima of a function,

Step 1. Find the first derivative of the function.

Step 2. Set the derivative equal to 0 now obtain the critical point.

Step 3. Evaluate f(x) at values of x to the left and one to the right of each critical point.

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.

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=16x6x4+7

Now, obtain its derivative as,

y=6x564x3=x54x3=x3(x24)

Now, set y=0 for critical values as,

x3(x24)=0

Thus,

x=0

And,

x=±2.

Now, substitute 0 for x in the provided equation to get,

y=00+7=7

And, substitute ±2 in the provided equation to get,

y=(±2)66(±2)4+7=53=1.67

Thus, the critical points are (0,7),(2,1.67) and (2,1.67).

Now, obtain the left and right behavior of these critical points.

Thus, substitute 1 for x in the derivative to get,

y=(1)54(1)3=1+4=3

Substitute 1 for x in the derivative as,

y=(1)54(1)3=14=3

Substitute 3 for x in the derivative to get,

y=(3)54(3)3=243108=135

Thus, y>0 for 2<x<0 and for x>2 and y<0 for x<2 and for 0<x<2.

Therefore, relative maximum occurs at (0,7) as derivative goes from positive to negative and relative minimum occurs at (2,53) and (2,53) as derivative goes from negative to positive.

Also, obtain the second derivative as,

y=x54x3y=5x412x2=x2(5x212)

Now, set y=0 to obtain the points of inflection as,

x2(5x212)=0

Thus, x=0 and x=±125 are points of inflection

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