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Math
Mathematical Applications for the Management, Life, and Social Sciences11th Edition

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Chapter 9.6, Problem 28E
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Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042

Solutions

Chapter
Section
Problem 1CP
Problem 2CP
Problem 1E
Problem 2E
Problem 3E
Problem 4E
Problem 5E
Problem 6E
Problem 7E
Problem 8E
Problem 9E
Problem 10E
Problem 11E
Problem 12E
Problem 13E
Problem 14E
Problem 15E
Problem 16E
Problem 17E
Problem 18E
Problem 19E
Problem 20E
Problem 21E
Problem 22E
Problem 23E
Problem 24E
Problem 25E
Problem 26E
Problem 27E
Problem 28E
Problem 29E
Problem 30E
Problem 31E
Problem 32E
Problem 33E
Problem 34E
Problem 35E
Problem 36E
Problem 37E
Problem 38E
Problem 39E
Problem 40E
Problem 41E
Problem 42E
Problem 43E
Problem 44E
Problem 45E
Problem 46E
Problem 47E
Problem 48E
Problem 49E
Problem 50E
Problem 51E
Problem 52E
Problem 53E
Problem 54E
Problem 55E
Problem 56E
BuyFindarrow_forward

Mathematical Applications for the ...

11th Edition
Ronald J. Harshbarger + 1 other
ISBN: 9781305108042
Textbook Problem

In Problems 25-28, write the equation of the line tangent to the graph of each function at the indicated point. As a check, graph both the function and the tangent line you found to see whether it looks correct.

26 .  y  =( x 2 +1) 3 at  ( 2, 125 )

To determine

To calculate: The equation of line tangent to the function y=(x2+1)3 at the point (2,125) and check by plotting the graph of the tangent line and the function.

Explanation

Given Information:

The function is y=(x2+1)3 and the line is tangent at the point (2,125).

Formula used:

Power rule:

If y=un and u is differentiable function of x, then,

dydx=nun−1⋅dudx

According to the property of differentiation, if a function is of the form f(x)=u(x)+v(x), then,

f′(x)=u′(x)+v′(x)

According to the product rule, if f(x)=u(x)â‹…v(x), then,

f′(x)=u′(x)⋅v(x)+v′(x)⋅u(x)

According to the point slope form, the equation of a line with slope m and point (x1,y1) is,

y−y1=m(x−x1)

Calculation:

Consider the provided function,

y=(x2+1)3

The equation of line tangent to the equation is found using the point-slope form.

The slope of tangent is found by differentiating the function.

Now, consider (x2+1) to be u,

y=u3

Differentiate both sides with respect to x,

dydx=ddx(u3)

Simplify using the power rule,

dydx=3⋅u3−1⋅dudx=3u2dudx

Substitute (x2+1) for u,

dydx=3(x2+1)2ddx(x2+1)=3(x2+1)2(ddx<

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