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(a) for what values of x is the function f(x) = |x² - 25| differentiabl…

Question

(a) for what values of x is the function f(x) = |x² - 25| differentiable? (enter your answer using interval notation.)
find a formula for f.
f(x) = { if |x| >
if |x| <
(b) sketch the graph of f.

Explanation:

Step1: Rewrite the function

First, rewrite \(y = |x^{2}-25|\) as a piece - wise function. We know that \(x^{2}-25=(x - 5)(x + 5)\). So \(y=

$$\begin{cases}x^{2}-25, & x\leq - 5\text{ or }x\geq5\\-(x^{2}-25),&-5

\)

Step2: Analyze differentiability

A function \(y = |u|\) is not differentiable at \(u = 0\). For \(u=x^{2}-25\), \(u = 0\) when \(x=-5\) and \(x = 5\). The function \(y = |x^{2}-25|\) is differentiable for all \(x
eq\pm5\). In interval notation, the domain of differentiability is \((-\infty,-5)\cup(-5,5)\cup(5,\infty)\)

Step3: Differentiate the piece - wise function

For \(|x|>5\) (i.e., \(x<-5\) or \(x > 5\)), \(f(x)=x^{2}-25\). Using the power rule \((x^{n})'=nx^{n - 1}\), \(f'(x)=2x\).
For \(|x|<5\) (i.e., \(-5

Answer:

\(x\in(-\infty,-5)\cup(-5,5)\cup(5,\infty)\)
\(f'(x)=

$$\begin{cases}2x, &\text{if }|x|>5\\-2x, &\text{if }|x|<5\end{cases}$$

\)

For part (b), to sketch the graph of \(y = |x^{2}-25|\):

  • First, consider the graph of \(y=x^{2}-25\), which is a parabola opening upwards with vertex at \((0, - 25)\) and \(x\) - intercepts at \(x=-5\) and \(x = 5\).
  • The graph of \(y = |x^{2}-25|\) takes the part of \(y=x^{2}-25\) that is above the \(x\) - axis (\(x\leq - 5\) or \(x\geq5\)) and reflects the part of \(y=x^{2}-25\) that is below the \(x\) - axis (\(-5