https://file.aiquickdraw.com/mathbot/file/1730581233219sh58ur5j.jpg

To tackle this problem, let's go through each part:

### Part (a)
1. **Determine Differentiability of \( f(x) = |x^2 - 9| \):**
   The function \( f(x) = |x^2 - 9| \) involves an absolute value, which can cause issues with differentiability at points where \( x^2 - 9 = 0 \). 

   Solving \( x^2 - 9 = 0 \):
   \[
   x^2 = 9 \implies x = \pm 3
   \]
   So, the points \( x = 3 \) and \( x = -3 \) may be points where \( f(x) \) is not differentiable, because the absolute value function \( |u| \) is not differentiable at \( u = 0 \).

2. **Express \( f(x) \) Piecewise:**
   Rewrite \( f(x) \) without the absolute value by considering cases for \( x^2 - 9 \):
   - When \( x > 3 \) or \( x < -3 \), \( x^2 - 9 > 0 \), so \( f(x) = x^2 - 9 \).
   - When \( -3 \leq x \leq 3 \), \( x^2 - 9 \leq 0 \), so \( f(x) = -(x^2 - 9) = 9 - x^2 \).

   Thus:
   \[
   f(x) = \begin{cases} 
      x^2 - 9 & \text{if } |x| > 3, \\
      9 - x^2 & \text{if } |x| \leq 3.
   \end{cases}
   \]

3. **Find \( f'(x) \):**
   Now, we can differentiate each piece separately:
   - For \( x > 3 \) or \( x < -3 \): \( f(x) = x^2 - 9 \Rightarrow f'(x) = 2x \).
   - For \( -3 < x < 3 \): \( f(x) = 9 - x^2 \Rightarrow f'(x) = -2x \).

   At \( x = \pm 3 \), there is a sharp corner due to the absolute value, so \( f(x) \) is **not differentiable** at \( x = 3 \) and \( x = -3 \).

   Therefore:
   \[
   f'(x) = \begin{cases} 
      2x & \text{if } |x| > 3, \\
      -2x & \text{if } |x| < 3.
   \end{cases}
   \]

### Part (b)
To sketch the graphs of \( f(x) \) and \( f'(x) \):

1. **Graph of \( f(x) \):**
   - For \( |x| > 3 \), plot \( f(x) = x^2 - 9 \), a parabola opening upwards shifted down by 9.
   - For \( |x| \leq 3 \), plot \( f(x) = 9 - x^2 \), a parabola opening downwards capped at \( x = \pm 3 \).

2. **Graph of \( f'(x) \):**
   - For \( |x| > 3 \), \( f'(x) = 2x \), a linear function with a positive slope.
   - For \( |x| < 3 \), \( f'(x) = -2x \), a linear function with a negative slope.
   - \( f'(x) \) has a discontinuity at \( x = \pm 3 \).

Would you like further details on any part, or assistance with the sketches?

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### Relative Questions
1. How does the absolute value affect differentiability in this function?
2. Why is \( f(x) \) not differentiable at \( x = \pm 3 \)?
3. What would \( f'(x) \) look like if there were no absolute value in \( f(x) \)?
4. How can we confirm differentiability of piecewise functions in general?
5. How does the derivative \( f'(x) \) represent changes in the graph of \( f(x) \)?

### Tip
When dealing with absolute value functions, remember to analyze the differentiability at points where the inside expression equals zero, as these points often lead to non-differentiable behavior.

For what values of x is the function f(x) = |x^2 - 9| differentiable? Find a formula for f'. Sketch the graphs of f and f'.

Explore the differentiability of the absolute value function f(x) = |x^2 - 9| and learn how to find its derivative, f'. Includes a detailed breakdown of differentiable intervals, piecewise function analysis, and sketching the graphs of f and f'. Suitable for undergraduate calculus students.

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