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Let

$ f(x) = \left\{

\begin{array}{ll}

x^2 + & \mbox{if} x \le 2\\

mx + b & \mbox{if} x > 2\\

\end{array} \right. $

Find the values of $ m $ and $ b $ that make $ f $ differentiable everywhere.

$$b=-4$$

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Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

It's close when you reign here, so for have to be differential everywhere. It has to be differential. X equals to us. Well, so the derivative one approaching from both sides of X equals two has to be equal to one another. Sorry. When we differentiate half a pecs is equal to two X and we get four. Well, we know that M X plus be is a line. So that means that the slope has to be equal to four since that day since the differentiated equation is equal to em for X, this bigger been too. And now we're going to find for be so f of two is equal to four if X is less than or equal to two. So we're gonna solve for B. Why is he equal to M X plus be? We plug in four equal to four times two plus B and we gotta be value of negative for