Question

(a) If \(g\left( x \right)={{x}^{{2}/{3}\;}}\)show that \(g'\left( 0 \right)\) does not exist.

(b) If \(a \ne 0\), find \(g'\left( a \right)\).

(c) Show that \(y={{x}^{{2}/{3}\;}}\)has a vertical tangent line at \(\left( {0,0} \right)\).

(d) Illustrate part (c) by graphing \(y={{x}^{{2}/{3}\;}}\)

Short answer

(a) It is proved that \(g'\left( 0 \right)\) does not exist.

(b) The value of \(g'\left( a \right)\) is \(\frac{2}{3{{a}^{{1}/{3}\;}}}\)

(c) It is proved that \(y = {x^{2/3}}\) has a vertical tangent line.

(d) The graph is shown below:

Step-by-step solution

The value of \(g'\left( 0 \right)\)

The value of \(g'\left( 0 \right)\) can be obtained as follows:

\begin{align}{g}'\left( 0 \right)&=\underset{x\to 0}{\mathop{\lim }}\,\frac{g\left( x \right)-g\left( 0 \right)}{x-0} \\ &=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{x}^{{2}/{3}\;}}-0}{x} \\ &=\underset{x\to 0}{\mathop{\lim }}\,\frac{1}{{{x}^{{1}/{3}\;}}} \\ &=\frac{1}{0} \\ &=\infty \end{align}

So, it is proved that \(g'\left( 0 \right)\) does not exist.

The derivative of the given function

(b)

The derivative of a function can be obtained using the formula given below:

\(g'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{g\left( x \right) - g\left( a \right)}}{{x - a}}\)

Substitute the value of \(g\left( x \right)\) in the above formula:

\(\begin{align}{g}'\left( a \right)&=\underset{x\to a}{\mathop{\lim }}\,\frac{g\left( x \right)-g\left( a \right)}{x-a} \\ &=\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{{2}/{3}\;}}-{{a}^{{2}/{3}\;}}}{x-a} \\ &=\underset{x\to a}{\mathop{\lim }}\,\frac{\left( {{x}^{{1}/{3}\;}}-{{a}^{{1}/{3}\;}} \right)\left( {{x}^{{1}/{3}\;}}+{{a}^{{1}/{3}\;}} \right)}{\left( {{x}^{{1}/{3}\;}}-{{a}^{{1}/{3}\;}} \right)\left( {{x}^{{2}/{3}\;}}+{{a}^{{2}/{3}\;}}+{{x}^{{1}/{3}\;}}{{a}^{{1}/{3}\;}} \right)} \\ &=\underset{x\to a}{\mathop{\lim }}\,\frac{{{x}^{{1}/{3}\;}}+{{a}^{{1}/{3}\;}}}{\left( {{x}^{{2}/{3}\;}}+{{a}^{{2}/{3}\;}}+{{x}^{{1}/{3}\;}}{{a}^{{1}/{3}\;}} \right)}\end{align}\)

Solve further the above equation,

\(\begin{align}{g}'\left( a \right)&=\frac{{{a}^{{1}/{3}\;}}+{{a}^{{1}/{3}\;}}}{\left( {{a}^{{2}/{3}\;}}+{{a}^{{2}/{3}\;}}+{{a}^{{1}/{3}\;}}{{a}^{{1}/{3}\;}} \right)} \\ &=\frac{2{{a}^{{1}/{3}\;}}}{3{{a}^{{2}/{3}\;}}} \\ &=\frac{2}{3{{a}^{{1}/{3}\;}}}\end{align}\)

Hence, the value of \(g'\left( a \right)\) is \(\frac{2}{3{{a}^{{1}/{3}\;}}}\)

The slope of the tangent

(c)

The slope of the tangent as \(x \to 0\) can be obtained as follows:

\(\begin{align}\underset{x\to 0}{\mathop{\lim }}\,\left| {g}'\left( x \right) \right|&=\underset{x\to 0}{\mathop{\lim }}\,\frac{2}{3{{x}^{{1}/{3}\;}}} \\ &=\frac{1}{0} \\ &=\infty \end{align}\)

Since \(g\) is continuous at \(x = 0\), since the slope of the tangent is not defined at \(x = 0\), so, \(g\) has a vertical tangent at \(x = 0\).

Check the answer visually

(d)

The procedure to draw the graph of the above equation by using the graphing calculator is as follows:

To check the answer, visually draw the graph of the function \(g\left( x \right)={{x}^{{2}/{3}\;}}\) by using the graphing calculator as shown below:

1. Open the graphing calculator. Select the “STAT PLOT” and enter the equation \({{x}^{{2}/{3}\;}}\)
2. Enter the “GRAPH” button in the graphing calculator.

Visualization of the graph of the function \(f\left( x \right)={{x}^{{2}/{3}\;}}\) is shown below:

In the graph, it is clear that the curve has a sharp corner at \(x = 0\). Therefore, the derivative does not exist at \(x = 0\) implies there exists vertical tangent at \(x = 0\).