Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 24

a.) Suppose that $\displaystyle G(x) = 4x^2 - x^3$, find $G'(a)$ and use it to find an equation of the tangent lines to the curve $\displaystyle y = 4x^2 - x^3$ at the point $(2,8)$ and $(3, 9)$

Using the definition of the derivative of a function $G$ at a number $a$, denoted by $G'(a)$, is

$\qquad \displaystyle \qquad G'(a) = \lim_{h \to 0} \frac{G(a + h) - G(a)}{h}$

We have,


$
\begin{equation}
\begin{aligned}

\qquad G'(a) =& \lim_{h \to 0} \frac{4(a + h)^2 - (a + h)^3 - (4a^2 - a^3)}{h}
&& \text{Substitute $G(a + h)$ and $G(a)$}\\
\\
\qquad G'(a) =& \lim_{h \to 0} \frac{4(a^2 + 2ah + h^2) - (a^3 + 3a^2 h + 3ah^2 + h^3) - 4a^2 + a^3}{h}
&& \text{Expand the equation}\\
\\
\qquad G'(a) =& \lim_{h \to 0} \frac{\cancel{4a^2} + 8ah + 4h^2 - \cancel{a^3} - 3a^2h - 3ah^2 - h^3 - \cancel{4a^2} + \cancel{a^3}}{h}
&& \text{Combine like terms}\\
\\
\qquad G'(a) =& \lim_{h \to 0} \frac{8ah + 4h^2 - 3a^2h - 3ah^2 - h^3}{h}
&& \text{Factor the numerator}\\
\\
\qquad G'(a) =& \lim_{h \to 0} \frac{\cancel{h} (8a + 4h - 3a^2 - 3ah - h^2)}{\cancel{h}}
&& \text{Cancel out like terms}\\
\\
\qquad G'(a) =& \lim_{h \to 0} (8a + 4h - 3a^2 - 5ah - h^2) = 8a + 4(0) - 3a^2 - 3a(0) - (0)^2
&& \text{Evaluate the limit}

\end{aligned}
\end{equation}
$


$\qquad \fbox{$G'(a) = 8a - 3a^2$} \qquad $ Equation of the slope of the tangent line

Solving for the slope and equation of the tangent line at $(2,8)$


$
\begin{equation}
\begin{aligned}

G'(2) =& 8(2) - 3(2)^2
&& \text{Substitute value of $a$}\\
\\
G'(2) =& 4
&& \text{Slope of the tangent line at $(2, 8)$}

\end{aligned}
\end{equation}
$


Using Point Slope Form where the tangent line $y = G(x)$ at $(a, G(a))$


$
\begin{equation}
\begin{aligned}

y - G(a) =& G'(a)(x - a)
&& \\
\\
y - 8 =& 4 (x - 2)
&& \text{Substitute value of $a, G(a)$ and $G'(a)$}\\
\\
y =& 4x - 8 + 8
&& \text{Combine like terms}

\end{aligned}
\end{equation}
$


$\qquad \qquad \fbox{$y = 4x$} \qquad $ Equation of the tangent line at $(2, 8)$

Solving for the slope and equation of the tangent line at $(3, 9)$


$
\begin{equation}
\begin{aligned}

G'(3) =& 8(3) - 3 (3)^2
&& \text{Substitute value of $(a)$}\\
\\

G'(3) =& -3
&& \text{Slope of the tangent line at $(3, 9)$}\\
\\
\end{aligned}
\end{equation}
$


Using Point Slope Form where the tangent line $y = G(x)$ at $(a, G(a))$


$
\begin{equation}
\begin{aligned}

y - 9 =& -3(x - 3)
&& \text{Substitute the value of the $a, G(a)$ and $G'(a)$}\\
\\
y =& -3x + 9 + 9
&& \text{Combine like terms}

\end{aligned}
\end{equation}
$


$\qquad\fbox{$ y =-3x + 18$} \qquad $ Equation of the tangent line at $(3, 9)$

b.) Draw a graph of the curve and the tangent lines on the same screen.

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