   Chapter 10.1, Problem 31E ### 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 31-36, both a function and its derivative are given. Use them to find critical values, critical points, intervals on which the function is increasing and decreasing, relative maxima, relative minima, and horizontal points of inflection; sketch the graph of each function. y = ( x 2 − 2 x ) 2 d y d x = 4 x ( x − 1 ) ( x − 2 )

To determine

To calculate: The critical values, critical points, the relative maxima, relative minima, or horizontal points of inflection of function y=(x22x)2 and its derivative dydx=4x(x1)(x2) and also sketch the graph.

Explanation

Given Information:

The provided function is y=(x22x)2 and the derivative is dydx=4x(x1)(x2).

Formula Used:

The simple power rule of derivative,

ddx(xn)=nxn1

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=(x22x)2,

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.

Now, equate the derivative to 0.

dydx=04x(x1)(x2)=0

From the property of multiplication,

4x=0x=0

Or,

x1=0x=1

Or,

x2=0x=2

Therefore, the critical values are 0, 1, and 2.

Substitute 0 for x in the function y=(x22x)2.

y=((0)22(0))2=00=0

Therefore, (0,0) is a critical point.

Substitute 1 for x in the function y=(x22x)2.

y=((1)22(1))2=(12)2=1

Therefore, (1,1) is a critical point.

Substitute 2 for x in the function y=(x22x)2.

y=((2)22(2))2=(44)2=0

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

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

Therefore the intervals to check the maxima, minima and point of inflection are (,0], [0,1], [1,2] and [2,).

For interval (,0]:

Substitute 1 for x in the function y=4x(x1)(x2)

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