## Calculus: Early Transcendentals 8th Edition

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