Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 48

a.) Suppose that $g(x) = \displaystyle x^{\frac{2}{3}}$, show that $g'(0)$ does not exist.

Using the definition of derivative


$\begin{equation} \begin{aligned} \qquad g'(x) =& \lim_{h \to 0} \frac{g(x + h) - g(x)}{h} && \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{(x + h)^{\frac{2}{3}} - (x)^{\frac{2}{3}}}{h} && \text{Substitute $g(x + h)$ and $g(x)$} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{(x + h)^{\frac{2}{3}} - (x)^{\frac{2}{3}}}{h} \cdot \frac{(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} + (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}}{(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} + (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}} && \text{Multiply both numerator and denominator by $[(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}]$. Recall that $(a^3 - b^3) = (a - b)(a^2 + ab + b^2)$} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{(x + h)^2 - \cancel{(x)^{\frac{2}{3}} (x +h)^{\frac{4}{3}}} + \cancel{(x)^{\frac{2}{3}} (x +h)^{\frac{4}{3}}} - \cancel{(x)^{\frac{4}{3}}(x + h)^{\frac{2}{3}}} + \cancel{(x)^{\frac{4}{3}}(x + h)^{\frac{2}{3}}} - x^2}{(h)[(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}]} && \text{Combine like terms} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{(x + h)^2 - x^2}{(h)[(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}]} && \text{Expand the equation} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{\cancel{x^2} + 2xh + h^2 - \cancel{x^2}}{(h)[(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}]} && \text{Combine like terms} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{2xh + h^2}{(h)[(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}]} && \text{Factor the numerator} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \frac{\cancel{h} (2x + h)}{\cancel{(h)}[(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}]} && \text{Cancel out like terms} \\ \\ \qquad g'(x) =& \lim_{h \to 0} \left[ \frac{2x + h}{(x + h)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + h)^{\frac{2}{3}} + (x)^{\frac{4}{3}}} \right] = \frac{2x + 0}{(x + 0)^{\frac{4}{3}} + (x)^{\frac{2}{3}} (x + 0)^{\frac{2}{3}} + (x)^{\frac{4}{3}}} = \frac{2x}{(x)^{\frac{4}{3}} + (x)^{\frac{2}{3}}(x)^{\frac{2}{3}} + (x)^{\frac{4}{3}}} = \frac{2x}{(x)^{\frac{4}{3}} + (x)^{\frac{4}{3}} + (x)^{\frac{4}{3}}} && \text{Evaluate the limit} \\ \\ \qquad g'(x) =& \frac{2x}{3(x)^{\frac{4}{3}}} && \text{Factor the denominator} \\ \\ \qquad g'(x) =& \frac{2 \cancel{x}}{3\cancel{(x)} (x)^{\frac{1}{3}}} && \text{Cancel out like terms} \\ \\ \qquad g'(x) =& \frac{2}{3(x)^{\frac{1}{3}}} && \text{Substitute $x$ which is zero} \\ \\ \qquad g'(0) =& \frac{2}{3(0)^{\frac{1}{3}}} && && \text{Simplify the equation} \\ \\ \qquad g'(0) =& \frac{2}{0} && \end{aligned} \end{equation}$


The function does not exist because the denominator is zero.

b.) Suppose that $a \neq 0$, find $g'(a)$.

Using the definition of derivative


$\begin{equation} \begin{aligned} \qquad f'(a) =& \lim_{x \to a} \frac{f(x) - f(a)}{x - a} && \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{(x)^{\frac{2}{3}} - (a)^{\frac{2}{3}}}{x - a} && \text{Substitute $f(x)$ and $f(a)$} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{ (x)^{\frac{2}{3}} - (a)^{\frac{2}{3}} }{x - a} \cdot \frac{ (x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}} }{(x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}}} && \text{Multiply both numerator and denominator by $[(x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}}]$. Recall that $(a^3 - b^3) = (a - b)(a^2 + ab + b^2)$} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{ x^2 - \cancel{(a)^{\frac{2}{3}} (x)^{\frac{4}{3}}} + \cancel{(a)^{\frac{2}{3}} (x)^{\frac{4}{3}}} - \cancel{(a)^{\frac{4}{3}} (x)^{\frac{2}{3}}} + \cancel{(a)^{\frac{4}{3}} (x)^{\frac{2}{3}}} - a^2 }{(x - a)[(x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}}]} && \text{Combine like terms} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{x^2 -a^2}{(x - a) [(x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}}]} && \text{Factor the numerator} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{\cancel{(x - a)} ( x + a)}{\cancel{(x - a)} [(x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}}]} && \text{Cancel out like terms} \\ \\ \qquad f'(a) =& \lim_{x \to a} \left[ \frac{x + a }{(x)^{\frac{4}{3}} + (ax)^{\frac{2}{3}} + (a)^{\frac{4}{3}}} \right] = \frac{a + a}{(a)^{\frac{4}{3}} + (a \cdot a)^{\frac{2}{3}} + (a)^{\frac{4}{3}}} = \frac{2a}{(a)^{\frac{4}{3}} + (a)^{\frac{4}{3}} +(a)^{\frac{4}{3}}} && \text{Evaluate the limit} \\ \\ f'(a) =& \frac{2a}{3(a)^{\frac{4}{3}}} && \text{Factor the denominator} \\ \\ f'(a) =& \frac{2\cancel{a}}{3\cancel{(a)}(a)^{\frac{1}{3}}} && \text{Cancel out like terms} \\ \\ f'(a) =& \frac{2}{3(a)^{\frac{1}{3}}} \text{ or } \frac{2}{3 \sqrt[3]{a}} \end{aligned} \end{equation}$


c.) Prove that $y = \displaystyle x^{\frac{2}{3}}$ has a vertical tangent line at $(0, 0)$

If the function has a vertical tangent line at $x = 0, \lim\limits_{x \to 0} f'(x) = \infty$

Given that, $f'(x) = \displaystyle \frac{2}{3(x)^{\frac{1}{3}}}$

Suppose that we substitute a value closer to 0 from the left and the right of the limit of $f'(x)$, let's say $x = -0.0000001$ and $x = 0.0000001$


$\begin{equation} \begin{aligned} & \lim_{x \to 0^-} \frac{2}{3 \sqrt[3]{-0.0000001}} = -143.629 \\ \\ & \lim_{x \to 0^+} \frac{2}{3 \sqrt[3]{0.0000001}} = 143.629 \end{aligned} \end{equation}$


This means that where $x$ gets closer and closer to from the left, the value of the limit approaches a very large negative number. On the other hand, as $x$ gets closer and closer to from the right, the value of the limit approaches a very large positive number. The tangent lines with these values become steeper and steeper as $x$ approaches until such time that the tangent line becomes a vertical line @ $x = 0$.