determining differentiability in exercises 77 - 80, describe the x - values at which f is differentiable.\n77…

Explanation:

Step1: Recall differentiability conditions

A function is differentiable at a point if it is continuous at that point and the left - hand and right - hand derivatives exist and are equal.

Step2: Analyze $f(x)=(x + 4)^{2/3}$

The derivative of $y=(x + 4)^{2/3}$ using the chain - rule $y^\prime=\frac{2}{3}(x + 4)^{-1/3}=\frac{2}{3\sqrt[3]{x + 4}}$. The function is not differentiable when the denominator is zero, i.e., when $x=-4$. So $f(x)$ is differentiable for all $x\neq - 4$.

Step3: Analyze $f(x)=\frac{x^{2}}{x^{2}-4}=\frac{x^{2}}{(x - 2)(x + 2)}$

The function is not defined at $x = 2$ and $x=-2$. Since differentiability implies continuity, $f(x)$ is not differentiable at $x = 2$ and $x=-2$. Also, the function has vertical asymptotes at these points. The derivative using the quotient - rule $y^\prime=\frac{2x(x^{2}-4)-x^{2}(2x)}{(x^{2}-4)^{2}}=\frac{2x^{3}-8x-2x^{3}}{(x^{2}-4)^{2}}=-\frac{8x}{(x^{2}-4)^{2}}$. So $f(x)$ is differentiable for all $x\neq\pm2$.

Step4: Analyze $f(x)=\sqrt{x + 1}+1$

The domain of the function is $x\geq - 1$. The derivative $y^\prime=\frac{1}{2\sqrt{x + 1}}$. The function is not differentiable at $x=-1$ (the left - hand derivative does not exist in the real - number system as we are dealing with the square - root function). So $f(x)$ is differentiable for all $x>-1$.

Step5: Analyze $f(x)=\begin{cases}x^{2}-4, &x\leq0\4 - x^{2},&x > 0\end{cases}$

First, find the left - hand derivative at $x = 0$ for $y=x^{2}-4$ ($x\leq0$), $y^\prime=2x$, and $y^\prime(0)=0$. Then find the right - hand derivative at $x = 0$ for $y = 4 - x^{2}$ ($x>0$), $y^\prime=-2x$, and $y^\prime(0)=0$. The function is continuous at $x = 0$ since $\lim_{x\rightarrow0^{-}}(x^{2}-4)=-4$ and $\lim_{x\rightarrow0^{+}}(4 - x^{2})=4$, and $f(0)=-4$. The function is differentiable for all real numbers $x$.

Answer:

• For $f(x)=(x + 4)^{2/3}$, $x\neq - 4$.
• For $f(x)=\frac{x^{2}}{x^{2}-4}$, $x\neq\pm2$.
• For $f(x)=\sqrt{x + 1}+1$, $x>-1$.
• For $f(x)=\begin{cases}x^{2}-4, &x\leq0\4 - x^{2},&x > 0\end{cases}$, all real $x$.