💬 👋 We’re always here. Join our Discord to connect with other students 24/7, any time, night or day.Join Here!

Like

Report

Numerade Educator

Like

Report

Problem 81 Medium Difficulty

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.

Answer

$$b=-4$$

More Answers

Discussion

You must be signed in to discuss.

Video Transcript

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