Suppose that f(x)=3√x, find f′(x),f″(x),f‴(x) and f^4(x). Graph f, f', f'' and f''' on a common screen. Are the graphs consistent with the geometric interpretations of these derivatives?
a.) Let a \neq 0, use the definition of derivative \displaystyle f'(a) = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x - a} to find f'(a).
Using the definition of derivative
\begin{equation} \begin{aligned} \qquad f'(a) =& \lim_{x \to a} \frac{\sqrt[3]{x} - \sqrt[3]{a}}{x - a} \cdot \frac{\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}}{\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}} && \text{Multiply both numerator and denominator by $\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}$} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{x + \cancel{\sqrt[3]{ax^2}} + \cancel{\sqrt[3]{a^2x}} - \cancel{\sqrt[3]{ax^2}} + \cancel{\sqrt[3]{a^2x}} - a}{(x - a)(\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2})} && \text{Combine like terms} \\ \\ \qquad f'(a) =& \lim_{x \to a} \frac{\cancel{x - a}}{\cancel{(x - a)}(\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}) } && \text{Cancel out like terms} \\ \\ \qquad f'(a) =& \lim_{x \to a} \left( \frac{1}{\sqrt[3]{x^2} + \sqrt[3]{ax} + \sqrt[3]{a^2}} \right) = \frac{1}{\sqrt[3]{a^2} + \sqrt[3]{(a)(a)} + \sqrt[3]{a^2}} && \text{Evaluate the limit} \\ \\ f'(a) =& \frac{1}{\sqrt[3]{a^2} + \sqrt[3]{a^2} + \sqrt[3]{a^2}} && \text{Combine like terms} \\ \\ f'(a) =& \frac{1}{3 \sqrt[3]{a^2}} \text{ or } \frac{1}{3(a)^{\frac{2}{3}}} && \end{aligned} \end{equation}
b.) Prove that f'(0) does not exist
Using f'(a) in part (a)
\begin{equation} \begin{aligned} f'(a) =& \frac{1}{3 \sqrt[3]{a^2}} \\ \\ f'(0) =& \frac{1}{3 \sqrt[3]{(0)^2}} \\ \\ f'(0) =& \frac{1}{3(0)} = \frac{1}{0} \end{aligned} \end{equation}
Therefore, f'(0) does not exist because denominator is zero.
c.) Prove that y = \sqrt[3]{x} has a vertical tangent line at (0,0)
If the function has a vertical tangent line at x = 0, \lim_{x \to 0} f'(x) = \infty
Given that f'(x) = \displaystyle \frac{1}{3 \sqrt[3]{x^2}}
Suppose that we substitute a value closer to from left and right to the limit of f'(x). Let's say x = -0.00001 and x = 0.000001
\begin{equation} \begin{aligned} & \lim_{x \to 0^-} \frac{1}{3\sqrt[3]{(-0.00001)^2}} = 2154.43 \\ \\ & \lim_{x \to 0^+} \frac{1}{3 \sqrt[3]{(0.0000001)^2}} = 46415.89 \end{aligned} \end{equation}
This means that where x gets closer and closer to , the value of limit approached a very large number. The tangent line with these values become steeper and steeper as x \to 0 until such time that the tangent line becomes a vertical line at x = 0.
Tuesday, May 28, 2013
Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 47
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