a graphing calculator is recommended. draw the graph of the function in a suitable viewing rectangle. y =…

Explanation:

Step1: Analyze domain

For the function $y = \log_{10}(4 - x^{2})$, the argument of the logarithm must be positive. So we set $4 - x^{2}>0$. Solving the inequality $(2 - x)(2 + x)>0$ gives $- 2<x<2$, so the domain is $(-2,2)$.

Step2: Analyze asymptotes

As $x\to - 2^{+}$ or $x\to2^{-}$, $4 - x^{2}\to0^{+}$, and $y=\log_{10}(4 - x^{2})\to-\infty$. So the vertical asymptotes are $x = - 2$ and $x = 2$.

Step3: Find derivative

First, use the chain - rule. If $y=\log_{10}(u)$ and $u = 4 - x^{2}$, then $y'=\frac{1}{u\ln(10)}\cdot(-2x)=\frac{-2x}{(4 - x^{2})\ln(10)}$.

Step4: Find critical points

Set $y' = 0$. Then $\frac{-2x}{(4 - x^{2})\ln(10)}=0$. The numerator $-2x = 0$ gives $x = 0$.

Step5: Analyze local extrema

When $x\in(-2,0)$, $y'>0$, the function is increasing. When $x\in(0,2)$, $y'<0$, the function is decreasing. So at $x = 0$, the function has a local maximum. Substitute $x = 0$ into $y=\log_{10}(4 - x^{2})$, we get $y=\log_{10}(4)\approx0.6$. There is no local minimum.

Answer:

domain: $(-2,2)$ asymptotes: $x=-2,x = 2$ local maximum: $(x,y)=(0,0.6)$ local minimum: DNE