Question

hw8 defining the derivative (targets l6, d3; §3.1)
score: 3/5 answered: 4/5
question 5
let ( f(x)=sqrt{67 - x}). compute ( f(3)) using the limit definition
( f(3)=)
find an equation of the tangent line at ( x = 3)
( y=)
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Explanation:

Step1: Recall limit - definition of derivative

The limit - definition of the derivative is $f^{\prime}(a)=\lim_{h
ightarrow0}\frac{f(a + h)-f(a)}{h}$. Here, $a = 3$ and $f(x)=\sqrt{67 - x}$, so $f(3)=\sqrt{67-3}=\sqrt{64} = 8$ and $f(3 + h)=\sqrt{67-(3 + h)}=\sqrt{64 - h}$.

Step2: Substitute into the limit - definition

$f^{\prime}(3)=\lim_{h
ightarrow0}\frac{\sqrt{64 - h}-8}{h}$. Multiply the numerator and denominator by the conjugate $\sqrt{64 - h}+8$:
\[

$$\begin{align*} f^{\prime}(3)&=\lim_{h ightarrow0}\frac{(\sqrt{64 - h}-8)(\sqrt{64 - h}+8)}{h(\sqrt{64 - h}+8)}\\ &=\lim_{h ightarrow0}\frac{(64 - h)-64}{h(\sqrt{64 - h}+8)}\\ &=\lim_{h ightarrow0}\frac{-h}{h(\sqrt{64 - h}+8)}\\ &=\lim_{h ightarrow0}\frac{-1}{\sqrt{64 - h}+8} \end{align*}$$

\]

Step3: Evaluate the limit

As $h
ightarrow0$, we have $f^{\prime}(3)=\frac{-1}{\sqrt{64}+8}=\frac{-1}{8 + 8}=-\frac{1}{16}$.

Step4: Find the equation of the tangent line

The equation of a tangent line is $y - y_0=m(x - x_0)$, where $(x_0,y_0)$ is the point of tangency and $m$ is the slope. Here, $x_0 = 3$, $y_0=f(3)=8$ and $m=f^{\prime}(3)=-\frac{1}{16}$.
\[

$$\begin{align*} y-8&=-\frac{1}{16}(x - 3)\\ y&=-\frac{1}{16}x+\frac{3}{16}+8\\ y&=-\frac{1}{16}x+\frac{3 + 128}{16}\\ y&=-\frac{1}{16}x+\frac{131}{16} \end{align*}$$

\]

Answer:

$f^{\prime}(3)=-\frac{1}{16}$
$y=-\frac{1}{16}x+\frac{131}{16}$