Single Variable Calculus, Chapter 3, 3.1, Section 3.1, Problem 18

a.) Determine the equation of the tangent line to the graph of $y=g(x)$ at $x = 5$ $\text{ if } g(5) = -3 \text{ and } g'(5) =4$




b.) Suppose that the tangent line to $y = f(x)$ at $(4,3)$ passes through the point $(0,2)$. Find $f(4)$ and $f'(4)$






$\text{a.) }$ Using point slope form


$y - y_1 = m(x-x_1)$

Recall from the definition that the derivative of the function is equal to the slope so,



$
\begin{equation}
\begin{aligned}
m &= g'(5) = 4\\
y-(-3) &= 4(x-5)\\
y+3 &= 4x-20\\
y &= 4x - 20 - 3\\
y &= 4x-23
\end{aligned}
\end{equation}
$


$\text{b.) }$ The line passes through the points $(4,3)$ and $(0,2)$ so we can use two
point form to determine the equation of the line.



$
\begin{equation}
\begin{aligned}
y - y_1 &= \frac{y_2-y_1}{x_2-x_1} (x-x_1)\\
y-3 &= \left( \frac{2-3}{0-4} \right) (x-4)\\
y-3 &= \frac{1}{4} (x-4)\\
y-3 &= \frac{x}{4}-1\\
y &= \frac{x}{4}+2\\
f(x) &= \frac{x}{4}+2\\
f(4) &= \frac{4}{4} +2 = 1+2 = 3\\
f(4) &= 3
\end{aligned}
\end{equation}
$



From the definition,



$
\begin{equation}
\begin{aligned}
f'(x) &= \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}\\
f'(x) &= \lim\limits_{h \to 0} \frac{\frac{x+h}{4}+2 - \left[\frac{x}{4}+2\right]}{h}\\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{\frac{x}{4}}+\frac{h}{4}+\cancel{2}-\cancel{\frac{x}{4}}+\cancel{2}}{h} \\
f'(x) &= \lim\limits_{h \to 0} \frac{\cancel{h}}{4\cancel{h}}\\
f'(x) &= \lim\limits_{h \to 0} \left(\frac{1}{4}\right)\\
f'(x) &= \frac{1}{4}\\
f'(4) &= \frac{1}{4}
\end{aligned}
\end{equation}
$

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