Saturday, November 30, 2013

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 @ $x = 0$.

No comments:

Post a Comment

Why is the fact that the Americans are helping the Russians important?

In the late author Tom Clancy’s first novel, The Hunt for Red October, the assistance rendered to the Russians by the United States is impor...