Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 16

Suppose that $f(x) = x^3$





a.) Estimate the values of $f'(0)$, $\displaystyle f'\left(\frac{1}{2}\right)$, $f'(1)$, $f'(2)$ and $f'(3)$
using the graph of $f$

b.) Use symmetry to deduce the values of $\displaystyle f'\left(-\frac{1}{2}\right)$, $f'(-1)$, $f(-2)$ and $f'(-3)$

c.) Use the values from parts(a) and (b) to graph $f'$

d.) Guess a formula for $f'(x)$

e.) Use the definition of a derivative to prove that your guess in part(d) is correct.



a.) Referring to the graph, $f'(0) \approx 0$, $\displaystyle f'\left(\frac{1}{2}\right) \approx 0.5$, $f'(1) \approx 4$,
$f'(2) \approx 11$ and $f'(3) \approx 25$

b.) By symmetry across the $x$-axis, it looks like the slopes are all the same or each sides of they $y$-axis
$\displaystyle f'\left(-\frac{1}{2}\right) \approx 0.5$, $f'(-1) \approx 4$, $f'(-2) \approx 11$ and $f'(-3) \approx 25$.

c.)



d.) Based from the symmetrical values of slopes across $y$-axis, we can form a formula for $f'(x)$ as $f'(x) = nx^2$; for $n > 0$
where $n$ could be any positive constant.


e.) Based from the definition of derivative,


$\quad \displaystyle f'(x) = \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} \qquad \text{ where } f(x) = x^3$


$
\begin{equation}
\begin{aligned}
f'(x) &= \lim\limits_{h \to 0} \frac{(x+3)^3-x^3}{h}\\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{x^3}+3x^2+h3xh^2+h^3-\cancel{x^3}}{h}\\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{h}(3x^2+3xh+h^2)}{\cancel{h}}\\
f'(x) &= \lim\limits_{h \to 0} (3x^2+3xh+h^2)\\
f'(x) &= 3x^2+3x(0) + (0)^2\\
f'(x) &= 3x^2
\end{aligned}
\end{equation}
$



It shows that part(d) and part(c) resembles each other.