Single Variable Calculus, Chapter 3, 3.3, Section 3.3, Problem 91

Where is the function $f(x) = |x^2-9|$ differentiable? Find a formula for $f'$ and sketch its graph.




Based from the graph, the values of $x$ from $-3 < x < 3$ are flipped or reflected to the $x$-axis.
By using the definition of the absolute value, we can deduce $f(x)$ as


$f(x) = \left\{ \begin{array}{c} x^2 - 9 & \text{for} & x \geq 3\\ 9-x^2 & \text{for} & -3 < x < 3\\ x^2 - 9 & \text{for} & x < -3 \end{array}\right.$


Now, we can find the formula for $f'(x)$ by taking the derivative of the Piecewise Function $f(x)$


$f'(x) = \left\{ \begin{array}{c} 2x & \text{for} & x \geq 3\\ -2x & \text{for} & -3 < x < 3\\ 2x & \text{for} & x \leq -3 \end{array}\right.$





Referring to the graph, we can say that $f(x)$ is differentiable every where except at $x = \pm 3$ because
of jump discontinuity making its limit from and right unequal.