Math Problem Statement
Solution
To tackle this problem, let's go through each part:
Part (a)
-
Determine Differentiability of : The function involves an absolute value, which can cause issues with differentiability at points where .
Solving : So, the points and may be points where is not differentiable, because the absolute value function is not differentiable at .
-
Express Piecewise: Rewrite without the absolute value by considering cases for :
- When or , , so .
- When , , so .
Thus:
x^2 - 9 & \text{if } |x| > 3, \\ 9 - x^2 & \text{if } |x| \leq 3. \end{cases}$$ -
Find : Now, we can differentiate each piece separately:
- For or : .
- For : .
At , there is a sharp corner due to the absolute value, so is not differentiable at and .
Therefore:
2x & \text{if } |x| > 3, \\ -2x & \text{if } |x| < 3. \end{cases}$$
Part (b)
To sketch the graphs of and :
-
Graph of :
- For , plot , a parabola opening upwards shifted down by 9.
- For , plot , a parabola opening downwards capped at .
-
Graph of :
- For , , a linear function with a positive slope.
- For , , a linear function with a negative slope.
- has a discontinuity at .
Would you like further details on any part, or assistance with the sketches?
Relative Questions
- How does the absolute value affect differentiability in this function?
- Why is not differentiable at ?
- What would look like if there were no absolute value in ?
- How can we confirm differentiability of piecewise functions in general?
- How does the derivative represent changes in the graph of ?
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