   Chapter 10.1, Problem 38E ### 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. f ( x ) = x 3 − 15 x 2 − 16 , 800 x + 80 , 000

To determine

To calculate: The location of critical points and determine a viewing window that shows by features of the graph f(x)=x315x216,800x+80,000.

Explanation

Given Information:

The provided function is f(x)=x315x216,800x+80,000.

Formula Used:

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.

This, in an interval between two critical values, the sign of the derivative at any value in the interval will be the sign of the derivative at all values in the interval.

As per the First Derivative Test,

The first derivative of the function is evaluated. The first derivative is made equal to zero in order to get the critical points.

The values of the critical values are kept inside the original function which gives the critical points. The intervals of the values of x are then evaluated for the relative maximum and minimum.

The viewing window is the interval that can be chosen from the critical points. It can differ for every person.

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 equation f(x)=x315x216,800x+80,000,

Use the simple power rule to differentiate,

dydx=3x230x16800=3(x210x+5600)=3(x+70)(x80)

Equate the above derivative to 0,

y=dydx=3(x+70)(x80)=0

Evaluate the values of x from the equation:

3(x+70)(x80)=0x=80,70

Hence, the value(s) of x are x=80,70.

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

Substitute 80 for x in the function f(x)=x315x216,800x+80,000.

y=(80)315(80)216,800(80)+80,000=512000960001344000+80000=848000

Hence, (80,848000) is a critical point.

Substitute 70 for x in the function f(x)=x315x216,800x+80,000

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