$$ f(x)=\begin{cases} x+b & \text{if } x \leq 1 \\ ax^2 & \text{if } x\gt 1 \end{cases} $$

Let \(f\) be the function above. What are all values of \(a\) and \(b\) for which \(f\) is differentiable at \(x=1\) ?

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We need to find values of \(a\) and \(b\) that make the following true:

$$ \lim\limits_{x\to 1^+} f'(x)=\lim\limits_{x\to 1^-} f'(x) $$

Differentiate each side of the piecewise function:

$$ f(x)=\begin{cases} x+b & \text{if } x \leq 1 \\ ax^2 & \text{if } x\gt 1 \end{cases} $$ $$ f'(x)=\begin{cases} 1 & \text{if } x \leq 1 \\ 2ax & \text{if } x\gt 1 \end{cases} $$ $$ 2ax=1 $$

When \(x=1\), $$ 2a(1)=1 $$ $$ a=\frac{1}{2} $$

Though we might be tempted to disregard the value of \(b\), we need to make sure the function is continuous at \(x=1\). For this we need:

$$ \lim\limits_{x\to 1^+} f(x)=\lim\limits_{x\to 1^-} f(x) $$ $$ ax^2=x+b $$

Using our values for \(a\) and \(x\):

$$ \left(\frac{1}{2}\right)(1)^2=1+b $$ $$ b=-\frac{1}{2} $$