Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises - Page 182: 81


$m=4$ and $b=-4$ is the answer to the question.

Work Step by Step

First, we would try to find $m$ and $b$ so that $f$ is differentiable at $x=2$. 1) For $f$ to be differentiable at $x=2$, first $f$ must be continuous at $x=2$ $\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^-}(x^2)=2^2=4$ $\lim\limits_{x\to2^+}f(x)=\lim\limits_{x\to2^+}(mx+b)=2m+b$ $f(2)=2^2=4$ For $f$ to be continuous at $x=2$, then $\lim\limits_{x\to2^-}f(x)=\lim\limits_{x\to2^+}=f(2)$. In other words, $$2m+b=4\hspace{1cm}(1)$$ 2) Then for $f$ to be differentiable everywhere, $f'(x)$ must exist everywhere. - For $x\le2$: $f'(x)=\frac{d}{dx}(x^2)=2x$ - For $x\gt2$: $f'(x)=\frac{d}{dx}(mx+b)=m$ Both of these derivatives all exist anywhere in its domain. However, for $f$ to be differentiable at $2$, both derivatives at $x=2$ must be equal. In other words, $$m=2\times2=4$$ Substitute $m=4$ into $(1)$, we have $$2\times4+b=4$$ $$8+b=4$$ $$b=-4$$ In conclusion, $m=4$ and $b=-4$ is the answer to the question.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.