Chapter 2.4, Problem 49E

To determine

To check: Whether the curve $y=x^{\frac{1}{3}}$ has a vertical tangent at $x=0$ or not.

Answer to Problem 49E

Yes, the curve $y=x^{\frac{1}{3}}$ has a vertical tangent at $x=0$.

Explanation of Solution

Given information: The curve is $y=x^{\frac{1}{3}}$ and the point is $x=0$.

Calculation:

The condition for the vertical tangent of $f\left(x\right)$ at $x=a$ is $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$ must tend to $\infty$ or $-\infty$.

The condition for the vertical tangent of $f\left(x\right)$ at $x=0$ is $\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$ must tend to $\infty$ or $-\infty$.

Substitute $h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^{\frac{1}{3}}\\ f\left(h\right)=h^{\frac{1}{3}}\end{array}$

Substitute $0$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=x^{\frac{1}{3}}\\ f\left(0\right)=0^{\frac{1}{3}}\\ f\left(0\right)=0\end{array}$

Substitute $h^{\frac{1}{3}}$ for $f\left(h\right)$ and $0$ for $f\left(0\right)$ in the formula.

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

Check the limit graphically.

To graph a function $y=h^{-\frac{2}{3}}$ , follow the steps using graphing calculator.

First press “ON” button on graphical calculator, press $\boxed{\text{Y}=}$ key and enter right hand side of the equation $y=h^{-\frac{2}{3}}$ after the symbol ${\text{Y}}_1$ . Enter the keystrokes $\boxed{(}\boxed{\text{X}}\boxed{)}\boxed{^}\boxed{(}\boxed{(-)}\boxed{2}\boxed{÷}\boxed{3}\boxed{)}$.

The display will show the equation,

  ${\text{Y}}_{\text{1}}=\left(\text{X}\right)^\left(-2/3\right)$

Now, press the $\boxed{\text{GRAPH}}$ key and $\boxed{\text{TRACE}}$ key to produce the graph of given function in standard window as shown below.

  

Figure (1)

As observed from the graph, the value of function tends to $\infty$ as $h$ tends to $0$ from the left side and the function tends to $\infty$ as $h$ tends to $0$ from the right side.

So, the right hand limit and left hand limit both tends to $\infty$ . The curve has vertical tangent.

Therefore, the curve $y=x^{\frac{1}{3}}$ has a vertical tangent at $x=0$.