Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 15

Suppose that $f(x) = x^2$



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



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



c.) Use the results from part(a) and (b) to guess a formula for $f'(x)$.



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






a.) Based from the graph, $f'(0) \approx 0$, $\displaystyle f'\left(\frac{1}{2}\right) \approx 1.25$, $f'(1) \approx 2$
and $f'(2) \approx 4.25$



b.) By symmetry across the $y$-axis, $\displaystyle f'\left( - \frac{1}{2}\right) \approx 1.25$, $f'(-1) \approx -2$ and
$f'(-2) \approx -4.25$



c.) Since the values of $f'(x)$ increases as $x$ gets bigger, we may form $f'(x) = nx$, where $n$ can be any positive integer.



d.) Based from the definition of the derivative.



$\quad \displaystyle f'(x) = \lim\limits_{h \to 0}\frac{f(x+h) - f(x)}{h} \qquad \text{ where } f(x) = x^2$



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



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