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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

In Problems 21-24, a function and its first and second derivatives are given. Use these to find relative maxima, relative minima, and points of inflection; sketch the graph of each function.

  f ( x )   = 3 x 5   20 x 3

f ( x ) = 15 x 2 ( x 2 )   ( x + 2 )

f " ( x )   = 60 x ( x 2     2 )

To determine

To calculate: The relative minimum, relative maximum and points of inflection for the provided function f(x)=3x520x3 and the first derivative f(x)=15x2(x2)(x+2) and second derivative f(x)=60x(x22) also sketch its graph.

Explanation

Given Information:

The provided function is f(x)=3x520x3, the first derivative is f(x)=15x2(x2)(x+2) and second derivative is f(x)=60x(x22).

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,

f(x)=3x520x3

Now, consider the first derivative f(x)=15x2(x2)(x+2).

Now, to obtain the critical values, set f(x)=0 as,

15x2(x2)(x+2)=0

Thus, either (x2)(x+2)=0 or 15x2=0.

First consider (x2)(x+2)=0.

Thus, x+2=0.

Subtract 2 from both sides as below,

x+22=02x=2

Now consider x2=0.

Add 2 on both sides as below,

x2+2=0+2x=2

And,

15x2=0x2=0x=0

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

Now, substitute 0 for x in the function f(x)=3x520x3,

f(x)=3(0)520(0)3=0

Substitute 2 for x in the function f(x)=3x520x3,

f(x)=3(2)520(2)3=96160=64

And, substitute 2 for x in the function f(x)=3x520x3,

f(x)=3(2)520(2)3=96+160=64

Thus, the critical points are (0,0), (2,64) and (2,64).

Now, consider the second derivative f(x)=60x(x22).

To obtain point of inflection, set f(x)=0 as,

60x(x22)=0

Thus, either x22=0 or 60x=0.

First consider x22=0.

Add 2 on both sides and take square root,

x22+2=0+2x2=2x=±2

And,

60x=0x=0

Now, substitute 0 for x in the function f(x)=3x520x3,

f(x)=3(0)520(0)3=0

Now, substitute 2 for x in the function f(x)=3x520x3,

f(x)=3x520x3=3(2)520(2)3=3(1

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