Chapter 2.7, Problem 50E

(a)

To determine

To Show: The function $g^{\prime }\left(0\right)$ does not exist.

Explanation of Solution

Result Used: The derivative of a function at $x=a$ is given by formula,

$g^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{g\left(x\right)-g\left(a\right)}{x-a}$ (1)

Proof:

Consider the function, $g\left(x\right)=x^{{}^{\frac{2}{3}}}$.

Compute $g^{\prime }\left(0\right)$ by using the equation (1),

$\begin{array}{c}{g}^{\prime }\left(0\right)=\underset{x\to 0}{\lim }\frac{g\left(x\right)-g\left(0\right)}{x-0}\\ =\underset{x\to 0}{\lim }\frac{x^{\frac{2}{3}}-{\left(0\right)}^{\frac{2}{3}}}{x}\\ =\underset{x\to 0}{\lim }\frac{x^{\frac{2}{3}}}{x}\\ =\underset{x\to 0}{\lim }\frac{x\cdot x^{-\frac{1}{3}}}{x}\end{array}$

$=\underset{x\to 0}{\lim }\frac{1}{x^{\frac{1}{3}}}$

Here, the function $\frac{1}{x^{\frac{1}{3}}}$ becomes larger and larger as x tends to zero. That is,

$\begin{array}{c}{g}^{\prime }\left(0\right)=\underset{x\to 0}{\lim }\frac{1}{x^{\frac{1}{3}}}\\ =\infty \end{array}$

Therefore, the derivative of the function does not exist at $x=0$.

Thus, the required proof is obtained.

(b)

To determine

To find: The value of $g^{\prime }\left(a\right)$ if $a\ne 0$.

Answer to Problem 50E

The value of derivative of $f\left(x\right)$ at $x=a$ is $\frac{2}{3a^{\frac{1}{3}}}$.

Explanation of Solution

Given:

The function $y=x^{\frac{2}{3}}$.

Calculation:

Obtain the derivative of the function $g\left(x\right)$ at $x=a$.

Compute $g^{\prime }\left(a\right)$ by using the equation (1),

$\begin{array}{c}{g}^{\prime }\left(x\right)=\underset{x\to a}{\lim }\frac{g\left(x\right)-g\left(a\right)}{x-a}\\ =\underset{x\to a}{\lim }\frac{x^{\frac{2}{3}}-a^{\frac{2}{3}}}{x-a}\\ =\underset{x\to a}{\lim }\frac{{\left(x^{\frac{1}{3}}\right)}^2-{\left(a^{\frac{1}{3}}\right)}^2}{{\left(x^{\frac{1}{3}}\right)}^3-{\left(a^{\frac{1}{3}}\right)}^3}\\ =\underset{x\to a}{\lim }\frac{\left(x^{\frac{1}{3}}+a^{\frac{1}{3}}\right)\left(x^{\frac{1}{3}}-a^{\frac{1}{3}}\right)}{\left(x^{\frac{1}{3}}-a^{\frac{1}{3}}\right)\left(x^{\frac{2}{3}}+a^{\frac{2}{3}}+x^{\frac{1}{3}}{a}^{\frac{1}{3}}\right)}\end{array}$$$

Simplify further,

$\begin{array}{c}{g}^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{\left(x^{\frac{1}{3}}+a^{\frac{1}{3}}\right)}{\left(x^{\frac{2}{3}}+a^{\frac{2}{3}}+x^{\frac{1}{3}}{a}^{\frac{1}{3}}\right)}\\ =\frac{\left({\left(a\right)}^{\frac{1}{3}}+a^{\frac{1}{3}}\right)}{\left({\left(a\right)}^{\frac{2}{3}}+a^{\frac{2}{3}}+{\left(a\right)}^{\frac{1}{3}}{a}^{\frac{1}{3}}\right)}\\ =\frac{2a^{\frac{1}{3}}}{3a^{\frac{2}{3}}}\\ =\frac{2}{3a^{\frac{1}{3}}}\end{array}$

Thus the value of the derivative at $x=a$ is, $g^{\prime }\left(a\right)=\frac{2}{3a^{\frac{1}{3}}}$.

(c)

To determine

To Show: The $y=x^{\frac{2}{3}}$ has a vertical tangent line at $\left(0,0\right)$.

Explanation of Solution

Result Used:

A curve has a vertical tangent line at $x=a$ if g is continuous at $x=a$ and $\underset{x\to a}{\lim }\left|g^{\prime }\left(x\right)\right|=\infty$

Proof:

Consider the equation $y=x^{\frac{2}{3}}$.

Substitute $x=0$ in $g\left(x\right)$,

$\begin{array}{c}g\left(0\right)={\left(0\right)}^{\frac{2}{3}}\\ =0\end{array}$

Thus $g\left(0\right)=0$ is defined.

The limit of the function $g\left(x\right)$ is computed as follows,

$\begin{array}{c}\underset{x\to 0}{\lim }g\left(x\right)=\underset{x\to 0}{\lim }{x}^{\frac{2}{3}}\\ ={\left(0\right)}^{\frac{2}{3}}\\ =0\end{array}$

Therefore, $g\left(x\right)$ is continuous at $x=0$.

From part (b), $g^{\prime }\left(x\right)=\frac{2x^{-\frac{1}{3}}}{3}$.

Take the limit of the function $g^{\prime }\left(x\right)$ as x approaches zero.

$\begin{array}{c}\underset{x\to 0}{\lim }\left|g^{\prime }\left(x\right)\right|=\underset{x\to 0}{\lim }\left|\frac{2x^{-\frac{1}{3}}}{3}\right|\\ =\frac{2}{3}\underset{x\to 0}{\lim }\left|x^{-\frac{1}{3}}\right|\\ =\frac{2}{3}\underset{x\to 0}{\lim }\left|\frac{1}{x^{\frac{1}{3}}}\right|\\ =\frac{2}{3}\underset{x\to 0}{\lim }\frac{1}{\left|x^{\frac{1}{3}}\right|}\end{array}$

$=\infty$

Since the function $g\left(x\right)$ is continuous at $x=0$ and $\underset{x\to 0}{\lim }\left|g^{\prime }\left(x\right)\right|=\infty$.

By result, the curve $g\left(x\right)$ has a vertical tangent at $x=0$.

Thus, the curve $g\left(x\right)$ has a vertical tangent at the point $\left(0,0\right)$.

(d)

To determine

To illustrate: Thepart (c) by graphing $y=x^{\frac{2}{3}}$.

Explanation of Solution

Graph:

Use the online graphing calculator to draw the graph of the function $y=x^{\frac{2}{3}}$ as shown below in Figure 1,

Illustration:

From Figure 1, it is clear that the y-axis is the vertical tangent to the curve $y=\sqrt[3]{x}$.