(a) for what values of x is the function f(x) = |x² - 25| differentiable? (enter your answer using interval…

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<x<5\end{cases})

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\neq\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<x<5)), (f(x)=-(x^{2}-25)=-x^{2}+25). Using the power rule, (f'(x)=-2x)

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<x<5)) above the (x) - axis. The graph has sharp corners at (x=-5) and (x = 5).