Question

use the limit definition of the derivative to find the slope of the tangent line to the function shown at x = 3. show your work. f(x)=\frac{3}{x}

Explanation:

Step1: Recall limit - definition of derivative

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

Step2: Substitute into the limit - definition

\[

$$\begin{align*} f^\prime(x)&=\lim_{h ightarrow0}\frac{\frac{3}{x + h}-\frac{3}{x}}{h}\\ &=\lim_{h ightarrow0}\frac{\frac{3x-3(x + h)}{x(x + h)}}{h}\\ &=\lim_{h ightarrow0}\frac{3x-3x-3h}{hx(x + h)}\\ &=\lim_{h ightarrow0}\frac{-3h}{hx(x + h)} \end{align*}$$

\]

Step3: Simplify the limit

Cancel out the $h$ terms: $f^\prime(x)=\lim_{h
ightarrow0}\frac{-3}{x(x + h)}$. As $h
ightarrow0$, we get $f^\prime(x)=-\frac{3}{x^{2}}$.

Step4: Evaluate the derivative at $x = 3$

Substitute $x = 3$ into $f^\prime(x)$: $f^\prime(3)=-\frac{3}{3^{2}}=-\frac{3}{9}=-\frac{1}{3}$.

Answer:

$-\frac{1}{3}$