Chapter 3.2, Problem 16E

To determine

Whether the given function $y$ is a corner, a cusp, a vertical tangent, or a discontinuity.

Answer to Problem 16E

The given function $y$ is a cusp.

Explanation of Solution

Given information:

The given function is $y=\sqrt[3]{\left|x\right|}$ and which is fails to be differentiable at $x=0$ .

Formula Used:

Differentiable function formula:

The differentiable function formulas to find the left hand derivative and right hand derivative are

Left hand derivative $=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ and Right hand derivative $=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ respectively.

Calculation:

The given function is:

  $y=\sqrt[3]{\left|x\right|}$

The given function can be written as:

  $\begin{array}{c}y=\{\begin{array}{l}\sqrt[3]{-x},x<0\\ \sqrt[3]{x},x\ge 0\end{array}\\ =\{\begin{array}{l}-\sqrt[3]{x},x<0\\ \sqrt[3]{x},x\ge 0\end{array}\end{array}$

To find the Left hand derivative, use the differentiable function formula. $\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Substitute $-\sqrt[3]{x}$ for $f\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Left}\text{hand}\text{derivative}=\underset{x\to 0^{-}}{\lim }\frac{-\sqrt[3]{x}-f\left(0\right)}{x-0}\\ =\underset{x\to 0^{-}}{\lim }\frac{-\sqrt[3]{x}-0}{x}\\ =\underset{x\to 0^{-}}{\lim }\left[-\frac{x^{1/3}}{x}\right]\\ =\underset{x\to 0^{-}}{\lim }\left[-\frac{1}{x^{2/3}}\right]\end{array}$

Simplify the above limit further.

  $\underset{x\to 0^{-}}{\lim }\left[-\frac{1}{x^{2/3}}\right]=-\infty$

Again, to find the Right hand derivative, use the differentiable formula $\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{f\left(x\right)-f\left(a\right)}{x-a}$ .

Substitute $\sqrt[3]{x}$ for $f\left(x\right)$ and $0$ for $a$ in the above formula.

  $\begin{array}{c}\text{Right}\text{hand}\text{derivative}=\underset{x\to 0^{+}}{\lim }\frac{\sqrt[3]{x}-f\left(0\right)}{x-0}\\ =\underset{x\to 0^{+}}{\lim }\frac{\sqrt[3]{x}-0}{x}\\ =\underset{x\to 0^{+}}{\lim }\left[\frac{x^{1/3}}{x}\right]\\ =\underset{x\to 0^{+}}{\lim }\left[\frac{1}{x^{2/3}}\right]\end{array}$

Simplify the above limit further.

  $\underset{x\to 0^{+}}{\lim }\left[\frac{1}{x^{2/3}}\right]=\infty$

Thus, the results of the limit indicate that the given function is a cusp.

The graph of the function $y=\sqrt[3]{\left|x\right|}$ has a cusp at $x=0$

  

Hence, the given function $y=\sqrt[3]{\left|x\right|}$ is a cusp.