   Chapter 9.5, Problem 25E ### 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 23-26, write the equation of the tangent line to the graph of the function at the indicated point. Check the reasonableness of your answer by graphing both the function and the tangent line. y = 3 x 4 − 2 x − 1 4 − x 2 at x = 1

To determine

To calculate: The equation of the tangent line to the graph of the function y=(3x42x1)4x2 at point x=1. Also, check the answer by graphing function and the tangent line.

Explanation

Given Information:

The provided function is y=(3x42x1)4x2.

Formula Used:

As per the quotient rule, if two functions are given in the form f(x)g(x), then the derivative is given as:

ddx(fg)=fggfg2

The slope (m) of the function is the first derivative of the function.

The equation of the tangent at any point (x1,y1) is given by

yy1=y(xx1)

Calculation:

Consider the provided function y=(3x42x1)4x2.

For the derivative of y=(3x42x1)4x2, follow the steps:

Consider f(x)=3x42x1 and g(x)=4x2.

Apply the quotient rule of the expression,

ddx(fg)=ddx(3x42x14x2)=(4x2)ddx(3x42x1)(3x42x1)ddx(4x2)(4x2)2

Evaluate the expression further,

ddx(fg)=(4x2)ddx(3x42x1)(3x42x1)ddx(4x2)(4x2)2=(4x2)(3ddx(x4)2ddx(x)ddx(1))(3x42x1)(ddx(4)ddx(x2))(4x2)2=(4x2)(12(x3)2)(3x42x1)(2x).(4x2)2=6x5+48x32x22x8(4x2)2

The slope of the tangent at x=1 can be found by the first derivative of the expression,

Hence the value of the slope is as follows:

y(1)=6(1)5+48(1)32(1)22(1)8(412)2=6+482289=103

Hence, the slope of the tangent line to the graph of the function y=(3x42x1)4x2 at x=1 is 103

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