Math Problem Statement

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'.

Solution

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

Part (a)

  1. Determine Differentiability of f(x)=x29f(x) = |x^2 - 9|: The function f(x)=x29f(x) = |x^2 - 9| involves an absolute value, which can cause issues with differentiability at points where x29=0x^2 - 9 = 0.

    Solving x29=0x^2 - 9 = 0: x2=9    x=±3x^2 = 9 \implies x = \pm 3 So, the points x=3x = 3 and x=3x = -3 may be points where f(x)f(x) is not differentiable, because the absolute value function u|u| is not differentiable at u=0u = 0.

  2. Express f(x)f(x) Piecewise: Rewrite f(x)f(x) without the absolute value by considering cases for x29x^2 - 9:

    • When x>3x > 3 or x<3x < -3, x29>0x^2 - 9 > 0, so f(x)=x29f(x) = x^2 - 9.
    • When 3x3-3 \leq x \leq 3, x290x^2 - 9 \leq 0, so f(x)=(x29)=9x2f(x) = -(x^2 - 9) = 9 - x^2.

    Thus:

    x^2 - 9 & \text{if } |x| > 3, \\ 9 - x^2 & \text{if } |x| \leq 3. \end{cases}$$
  3. Find f(x)f'(x): Now, we can differentiate each piece separately:

    • For x>3x > 3 or x<3x < -3: f(x)=x29f(x)=2xf(x) = x^2 - 9 \Rightarrow f'(x) = 2x.
    • For 3<x<3-3 < x < 3: f(x)=9x2f(x)=2xf(x) = 9 - x^2 \Rightarrow f'(x) = -2x.

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

    Therefore:

    2x & \text{if } |x| > 3, \\ -2x & \text{if } |x| < 3. \end{cases}$$

Part (b)

To sketch the graphs of f(x)f(x) and f(x)f'(x):

  1. Graph of f(x)f(x):

    • For x>3|x| > 3, plot f(x)=x29f(x) = x^2 - 9, a parabola opening upwards shifted down by 9.
    • For x3|x| \leq 3, plot f(x)=9x2f(x) = 9 - x^2, a parabola opening downwards capped at x=±3x = \pm 3.
  2. Graph of f(x)f'(x):

    • For x>3|x| > 3, f(x)=2xf'(x) = 2x, a linear function with a positive slope.
    • For x<3|x| < 3, f(x)=2xf'(x) = -2x, a linear function with a negative slope.
    • f(x)f'(x) has a discontinuity at x=±3x = \pm 3.

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


Relative Questions

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

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Math Problem Analysis

Mathematical Concepts

Differentiability
Absolute Value Functions
Piecewise Functions
Derivative

Formulas

f(x) = |x^2 - 9|
f'(x) = 2x for |x| > 3
f'(x) = -2x for |x| < 3

Theorems

Differentiability and continuity of absolute value functions
Piecewise differentiability

Suitable Grade Level

Undergraduate Calculus