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 @ $x = 0$.
Saturday, November 30, 2013
Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 48
Subscribe to:
Post Comments (Atom)
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...
-
There are a plethora of rules that Jonas and the other citizens must follow. Again, page numbers will vary given the edition of the book tha...
-
The poem contrasts the nighttime, imaginative world of a child with his daytime, prosaic world. In the first stanza, the child, on going to ...
-
The given two points of the exponential function are (2,24) and (3,144). To determine the exponential function y=ab^x plug-in the given x an...
-
The play Duchess of Malfi is named after the character and real life historical tragic figure of Duchess of Malfi who was the regent of the ...
-
The only example of simile in "The Lottery"—and a particularly weak one at that—is when Mrs. Hutchinson taps Mrs. Delacroix on the...
-
Hello! This expression is already a sum of two numbers, sin(32) and sin(54). Probably you want or express it as a product, or as an expressi...
-
Macbeth is reflecting on the Weird Sisters' prophecy and its astonishing accuracy. The witches were totally correct in predicting that M...
No comments:
Post a Comment