Question

Find the values of m and b that make the following function differentiable.

Answer 1

$f(x)=\begin{cases}x^2&\text{for }x\le2\\mx+b&\text{for }x>2\end{cases}$

In order to be differentiable everywhere, $f$ must first be continuous everywhere, which means the limits from either side as $x\to2$ must be the same and equal to $f(2)$. By definition, $f(2)=2^2=4$, and

$\displaystyle\lim_{x\to2^-}f(x)=\lim_{x\to2}x^2=4$

$\displaystyle\lim_{x\to2^+}f(x)=\lim_{x\to2}(mx+b)=2m+b$

so we need to have $4m+b=4$.

For $f$ to be differentiable at $x=2$, the derivative needs to be continuous at $x=2$, i.e.

$\displaystyle\lim_{x\to2^-}f'(x)=\lim_{x\to2^+}f'(x)$

We then need to have

$\displaystyle\lim_{x\to2}2x=\lim_{x\to2}m\implies\boxed{m=4}$

Then

$2m+b=4\implies\boxed{b=-4}$