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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

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BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

Find the relative maxima, relative minima, and points of inflection and sketch the graphs of the functions in Problems 15-20.

y   =   x 4 8 x 3 +   16 x 2

To determine

To calculate: The relative minimum, relative maximum and points of inflection for the provided function y=x48x3+16x2 and sketch its graph.

Explanation

Given Information:

The provided function is,

y=x48x3+16x2

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 function,

y=x48x3+16x2

Now, calculate the first derivative.

y=x48x3+16x2y=4x324x2+32x=4x(x26x+8)

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

4x(x26x+8)=0

Thus, either x26x+8=0 or 4x=0.

First consider x26x+8=0.

Factorize x26x+8.

x26x+8=(x4)(x2)

Thus, (x4)(x2)=0.

First consider x4=0.

Add 4 on both sides.

x4+4=0+4x=4

Now consider x2=0.

Add 2 on both sides.

x2+2=0+2x=2

And,

4x=0x=0

Thus, the critical values of the function are at x=2, x=0 and x=4.

Now, substitute 0 for x in the equation y=x48x3+16x2,

y=048(0)3+16(0)2=0

Substitute 2 for x in the equation y=x48x3+16x2,

y=248(2)3+16(2)2=1664+64=16

And, substitute 4 for x in the equation y=x48x3+16x2,

y=448(4)3+16(4)2=256512+256=0

Thus, the critical points are (0,0), (2,16) and (4,0).

Now, obtain the second derivative as,

y=12x248x+32=4(3x212x+8)

To obtain point of inflection, set y=0 as,

4(3x212x+8)=03x212x+8=0

Use discriminant method x=b±b24ac2a to solve 3x212x+8 for x

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Chapter 10 Solutions

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