An example of a function that is continuous at $x = -2$, $x = 0$, and $x = 3$ but whose derivative does not exist at these points is the absolute value function combined with linear shifts:

$f(x) = |x + 2| + |x| + |x - 3|$

Explanation:

1. Continuity at the points:

   • The absolute value function $|x|$ is continuous for all $x$.
   • Therefore, $f(x)$ is continuous at $x = -2$, $x = 0$, and $x = 3$.

2. Derivative at the points:

   • The derivative of $|x|$ is not defined at $x = 0$ because the slope of the function changes abruptly at this point (it goes from $-1$ to $1$).
   • Similarly, the derivative of $|x + 2|$ is not defined at $x = -2$ and the derivative of $|x - 3|$ is not defined at $x = 3$ for the same reason.

Thus, $f(x)$ is continuous at $x = -2$, $x = 0$, and $x = 3$, but its derivative does not exist at these points.

Would you like further details on how the derivative behaves at these points or have any other questions?

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Here are 5 related questions:

1. How do you formally prove the continuity of a piecewise function?
2. What are other types of points where a function can be continuous but not differentiable?
3. Can you provide a function that is not continuous at some points but has a derivative at other points?
4. What is the difference between a cusp and a corner in a graph, and how does it affect differentiability?
5. How does the absolute value function behave differently from polynomial functions in terms of continuity and differentiability?

Tip: The derivative may fail to exist at points where the function has a sharp corner, a cusp, or a vertical tangent.