graphical analysis in exercises 81 - 84, use a graphing utility to graph the function and find the x…

Explanation:

Step1: Recall differentiability conditions

A function is not differentiable at points of sharp - corners, vertical tangents, or discontinuities.

Step2: Analyze (f(x)=|x - 5|)

Rewrite (f(x)) as a piece - wise function: (f(x)=\begin{cases}x - 5, &x\geq5\-(x - 5),&x<5\end{cases}). The graph has a sharp corner at (x = 5). So (f(x)) is differentiable for all (x\neq5).

Step3: Analyze (f(x)=\frac{4x}{x - 3})

The function (f(x)=\frac{4x}{x - 3}) has a vertical asymptote at (x = 3) (since the denominator is zero at (x = 3)). It is differentiable for all (x\neq3).

Step4: Analyze (f(x)=x^{\frac{2}{5}})

Find the derivative using the power rule (y = x^n\Rightarrow y^\prime=nx^{n - 1}), (f^\prime(x)=\frac{2}{5}x^{-\frac{3}{5}}=\frac{2}{5x^{\frac{3}{5}}}). The function has a vertical tangent at (x = 0). So it is differentiable for all (x\neq0).

Step5: Analyze (f(x)=\begin{cases}x^{3}-3x^{2}+3x, &x\leq1\x^{2}-2x, &x > 1\end{cases})

First, find the derivative of (y_1=x^{3}-3x^{2}+3x), (y_1^\prime=3x^{2}-6x + 3). Evaluate at (x = 1), (y_1^\prime(1)=3-6 + 3=0). Then, find the derivative of (y_2=x^{2}-2x), (y_2^\prime=2x-2). Evaluate at (x = 1), (y_2^\prime(1)=2 - 2=0). The function is differentiable for all real (x).

Answer:

For (f(x)=|x - 5|), (x\in(-\infty,5)\cup(5,\infty)); for (f(x)=\frac{4x}{x - 3}), (x\in(-\infty,3)\cup(3,\infty)); for (f(x)=x^{\frac{2}{5}}), (x\in(-\infty,0)\cup(0,\infty)); for (f(x)=\begin{cases}x^{3}-3x^{2}+3x, &x\leq1\x^{2}-2x, &x > 1\end{cases}), (x\in(-\infty,\infty))