Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 29

a.) Suppose that $f(x) = x^4 + 2x$, find $f(x)$

Using the definition of derivative




$
\begin{equation}
\begin{aligned}

\qquad f'(x) =& \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
&&
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{(x + h)^4 + 2 (x + h) - (x^4 + 2x)}{h}
&& \text{Substitute $f(x + h)$ and $f(x)$}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{x^4 + 4x^3 + 6x^2 h^2 + 4xh^3 + h^4 + 2x + 2h - x^4 - 2x}{h}
&& \text{Expand the equation}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{x^4} + 4x^3 + 6x^2 h^2 + 4xh^3 + h^4 + \cancel{2x} + 2h - \cancel{x^4} - \cancel{2x}}{h}
&& \text{Combine like terms}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{4x^3h + 6x^2 h^2 + 4xh^3 + h^4 + 2h}{h}
&& \text{Factor the numerator}
\\
\\
\qquad f'(x) =& \lim_{h \to 0} \frac{\cancel{h}(4x^3 + 6x^2 h + 4xh^2 + h^3 + 2)}{\cancel{h}}
&& \text{Cancel out like terms}
\\
\\
f'(x) =& \lim_{h \to 0} (4x^3 + 6x^2 h + 4xh^2 + h^3 + 2)
&& \text{Evaluate the limit}
&&
\\
\\
f'(x) =& 4x^3 + 6x^2 (0) + 4x(0)^2 + (0)^3 + 2 \quad = 4x^3 + 0 + 0 + 0 + 2 \\
\\

f'(x) =& 4x^3 + 2
&&

\end{aligned}
\end{equation}
$


b.) Compare the graphs of $f$ and $f'$ and check whether your answer in part (a) is reasonable.