Chapter 2.7, Problem 49E

(a)

To determine

To find: The derivative of the function $f\left(x\right)$ at $x=a$ if $a\ne 0$.

Answer to Problem 49E

The value of $f^{\prime }\left(a\right)$ is $\frac{1}{3}{a}^{-\frac{2}{3}}$.

Explanation of Solution

Given:

The given function is, $f\left(x\right)=\sqrt[3]{x}$.

Result Used:

The derivative of a function f at $x=a$, denoted by $f^{\prime }\left(x\right)$, is

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

Difference of cube formula: $x^3-y^3=\left(x-a\right)\left(x^2+y^2+xy\right)$.

Calculation:

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

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

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

Apply the difference of cube formula in the numerator as follows,

$\begin{array}{c}{f}^{\prime }\left(a\right)=\underset{x\to a}{\lim }\frac{{\left(x^{\frac{1}{3}}\right)}^3-{\left(a^{\frac{1}{3}}\right)}^3}{x-a}\times \frac{1}{\left(x^{\frac{2}{3}}+a^{\frac{1}{3}}{x}^{\frac{1}{3}}+a^{\frac{2}{3}}\right)}\\ =\underset{x\to a}{\lim }\frac{x-a}{x-a}\times \frac{1}{\left(x^{\frac{2}{3}}+a^{\frac{1}{3}}{x}^{\frac{1}{3}}+a^{\frac{2}{3}}\right)}\\ =\underset{x\to a}{\lim }\frac{1}{\left\{x^{\frac{2}{3}}+a^{\frac{1}{3}}{x}^{\frac{1}{3}}+a^{\frac{2}{3}}\right\}}\left[\because x-a\ne 0\right]\\ =\frac{\underset{x\to a}{\lim }\left(1\right)}{\underset{x\to a}{\lim }\left\{x^{\frac{2}{3}}+a^{\frac{1}{3}}{x}^{\frac{1}{3}}+a^{\frac{2}{3}}\right\}}\end{array}$

$=\frac{1}{a^{\frac{2}{3}}+a^{\frac{1}{3}}{a}^{\frac{1}{3}}+a^{\frac{2}{3}}}$

Simplify the denominator,

$\begin{array}{c}{f}^{\prime }\left(a\right)=\frac{1}{a^{\frac{2}{3}}+a^{\frac{2}{3}}+a^{\frac{2}{3}}}\\ =\frac{1}{3a^{\frac{2}{3}}}\\ =\frac{1}{3}{a}^{-\frac{2}{3}}\end{array}$

Thus, the value of $f^{\prime }\left(a\right)$ is $\frac{1}{3}{a}^{-\frac{2}{3}}$.

(b)

To determine

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

Explanation of Solution

Result used:

The derivative of a function f , denoted by $f^{\prime }\left(x\right)$, is

$f^{\prime }\left(a\right)=\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ (2)

Proof:

Consider the function $f\left(x\right)=\sqrt[3]{x}$.

Compute $f^{\prime }\left(0\right)$ by using the equation (2),

$\begin{array}{c}{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{\sqrt[3]{0+h}-\sqrt[3]{0}}{h}\\ =\underset{h\to 0}{\lim }\frac{\sqrt[3]{h}-0}{h}\\ =\underset{h\to 0}{\lim }\frac{h^{\frac{1}{3}}}{h}\\ =\underset{h\to 0}{\lim }\frac{1}{h\cdot h^{-\frac{1}{3}}}\end{array}$

$=\underset{h\to 0}{\lim }\frac{1}{h^{\frac{2}{3}}}$

Here, the function $\frac{1}{h^{\frac{2}{3}}}$ becomes larger and larger as h tends to zero. That is,

$\begin{array}{c}{f}^{\prime }\left(0\right)=\underset{h\to 0}{\lim }\frac{1}{h^{\frac{2}{3}}}\\ =\infty \end{array}$

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

Thus, the required proof is obtained.

(c)

To determine

To show: The $y=\sqrt[3]{x}$ 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 f is continuous at $x=a$ and $\underset{x\to a}{\lim }\left|f^{\prime }\left(x\right)\right|=\infty$

Proof:

Consider the equation $f\left(x\right)=\sqrt[3]{x}$.

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

$\begin{array}{c}f\left(0\right)=\sqrt[3]{0}\\ =0\end{array}$

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

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

$\begin{array}{c}\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\sqrt[3]{x}\\ =\sqrt[3]{0}\\ =0\end{array}$

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

From part (a), $f^{\prime }\left(x\right)=\frac{x^{-\frac{2}{3}}}{3}$

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

$\begin{array}{c}\underset{x\to 0}{\lim }\left|f^{\prime }\left(x\right)\right|=\underset{x\to 0}{\lim }\left|\frac{1}{3x^{\frac{2}{3}}}\right|\\ =\frac{1}{3}\underset{x\to 0}{\lim }\left|\frac{1}{x^{\frac{2}{3}}}\right|\\ =\frac{1}{3}\underset{x\to 0}{\lim }\frac{1}{x^{\frac{2}{3}}}\\ =\infty \left[\text{From part (b)}\right]\end{array}$

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

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

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

Graph:

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

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