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Chapter 2: Q. 94 (page 186)

Use Problem 93 to prove that a linear function is its own tangent line at every point. In other words, show that if f(x)=mx+bis any linear function, then the tangent line tofat any point x=cis given by y=mx+b.

Short Answer

Expert verified

Ans:

y-f(c)=f'(c)[x-c]y-(mc+b)=m[x-c]y-mc-b=mx-mcy=mx-mc+mc+by=mx+b

Step by step solution

01

Step 1. Given information:

Consider the function:

f(x)=mx+b

Here the objective is to show that the equation of the tangent line at any point x=cis y=mx+b.

02

Step 2. Solving the equation:

f'(x)=limz鈫�4f(z)-f(x)z-x=limz鈫�x(mz+b)-(mx+b)z-x=limz鈫�xm(z-x)z-x=limz鈫�xm=m

Hence the slope of the linear function is constant and it is f'(x)=m

At any point x=c the slope of the function is f'(c)=m

At x=c, the value of the function is

f(c)=mc+b

03

Step 3. Finding the equation of the tangent line:

The eq of the tangent line will be,

y-f(c)=f'(c)[x-c]y-(mc+b)=m[x-c]y-mc-b=mx-mcy=mx-mc+mc+by=mx+b

Hence the eq of the tangent is same as linear function.

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Most popular questions from this chapter

Find the derivatives of the functions in Exercises 21鈥�46. Keep in mind that it may be convenient to do some preliminary algebra before differentiating.

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