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

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 $x$:

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