Calculus and Its Applications, Chapter 1, 1.3, Section 1.3, Problem 48

Determine the simplified difference quotient of the function $f(x) = ax^3 + bx^2$
For $f(x) = ax^3 + bx^2$

$
\begin{equation}
\begin{aligned}
f(x + h) &= a(x + h)^3 + b(x +h)^2\\
\\
&= a \left[ x^3 + 3x^2h + 3xh^2 + h^3 \right] + b\left[ x^2 + 2xh + h^2 \right]\\
\\
&= ax^3 + 3ax^2h + 3axh^2 + ah^3 + bx^2 + 2bxh + bh^2
\end{aligned}
\end{equation}
$

Then,

$
\begin{equation}
\begin{aligned}
f(x +h) - f(x) &= ax^3 + 3ax^2h + 3axh^2 + ah^3 + bx^2 + 2bxh + bh^2 - \left( ax^3 + bx^2 \right)\\
\\
&= ax^3 + 3ax^2h + 3axh^2 + ah^3 + bx^2 + 2bxh + bh^2 - ax^3 - bx^2\\
\\
&= 3ax^2h + 3axh^2 + ah^3 + 2bxh + bh^2
\end{aligned}
\end{equation}
$


Thus,

$
\begin{equation}
\begin{aligned}
\frac{f(x+h)-f(x)}{h} &= \frac{3ax^2h + 3axh^2 + ah^3 + 2bxh + bh^2}{h}\\
\\
&= \frac{h\left( 3ax^2 + 3axh + ah^2 + 26x + bh \right)}{h} \\
\\
&= 3ax^2 + 3axh + ah^2 + 2bx + bh
\end{aligned}
\end{equation}
$

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