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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 1-4, find all critical points and determine whether they are relative maxima, relative minima, or horizontal points of inflection.

f ( x ) = 1 3 x + 3 x 2 x 3

To determine

To calculate: The critical values, relative maxima, relative minima, or horizontal points of inflection of the function f(x)=13x+3x2x3 by using the derivative.

Explanation

Given Information:

The provided equation is f(x)=13x+3x2x3.

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.

Calculation:

Consider the provided equation f(x)=13x+3x2x3,

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.

Take out the first derivative of the equation by the power rule,

f(x)=3+6x3x2

Put the value of y=0,

f(x)=3+6x3x2=0=3(12x+x2)=0=12x+x2=0

Hence, solve the quadratic equation:

12x+x2=01xx+x2=0(1x)x(1x)=0

Hence, the values of x is x=1

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