Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 90

Where is the function
$
f(x) = \left\{
\begin{array}{c}
-1-2x & \text{if} & x < -1\\
x^2 & \text{if} & -1 \leq x \leq 1\\
x & \text{if} & x > 1
\end{array}\right.
$
differentiable? Give a formula for $f'$ and sketch the graph of $f$ and $f'$


In order for us to know what numbers is function $f$ differentiable, we must prove
that the function is continuous at that number, that is $f'_- (x)$ and $f'_+ (x)$ should be equal.

For $x=-1$

$
\begin{equation}
\begin{aligned}
f_-(-1) & = -1-2x\\
f'_- (-1) &= -1-2\frac{d}{dx}(x) \quad = -2(1) \quad = -2\\
f_+ (-1) &= x^2\\
f'_+ (-1) &= \frac{d}{dx}(x^2) \quad = 2x \quad = 2(-1) \quad = -2
\end{aligned}
\end{equation}
$


For $x=1$

$
\begin{equation}
\begin{aligned}
f_- (1) &= x^2\\
f'_-(1) &= \frac{d}{dx}(x^2) \quad = 2x \quad 2(1) \quad =2\\
f_+ (1) &= x\\
f'_+(1) &= \frac{d}{dx}(x) \quad =x
\end{aligned}
\end{equation}
$


We can find the formula for $f'(x)$ by taking the derivative of $f(x)$.
Therefore, the formula for $f'(x)$ is
$
f'(x) = \left\{
\begin{array}{c}
-2 & \text{if} & x < -1\\
2x & \text{if} & -1 \leq x \leq 1\\
1 & \text{if} & x >1
\end{array}\right.
$