   Chapter 10.1, Problem 42E ### 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 37-42, use the derivative to locate critical points and determine a viewing window that shows all features of the graph. Use a graphing calculator to sketch a complete graph. y = 7 + 3 x 5 − 5 x 6

To determine

To calculate: The location of critical points from the derivation of the function f(x)=7+3x55x6 and determine viewing window that shows all features of the graph of f(x)=7+3x55x6 and sketch the complete graph.

Explanation

Given Information:

The provided function is f(x)=7+3x55x6.

Formula Used:

The simple power rule to derivative.

ddx(xn)=nxn1

The steps to calculate the critical values of a function

Step 1: First find the derivative of the function.

Step 2: Equate the derivate of the function to 0 and calculate the possible critical points.

Step 3: Put the critical point in the provided function to get the critical values.

Calculation:

Consider the provided function f(x)=7+3x55x6,

Use the simple power rule to differentiate,

dydx=15x430x5=15x4(12x)

Equate the above derivative to 0,

y=015x4(12x)=0

Evaluate the values of x from the equation:

15x4(12x)=0

From the property of zero multiplication,

15x4=0x=0

Or,

12x=02x=1 x=0.5

Hence, the critical values of x are x=0 and 0.5.

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

Substitute 0 for x in f(x)=7+3x55x6.

y=7+3(0)55(0)6=7+00=7

Hence, (0,7) is a critical point.

Substitute 0.5 for x in f(x)=7+3x55x6.

y=7+3(0.5)55(0.5)6=7+0.031250.078125=7.015625

Hence, (0

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