Question

homework4: problem 3 (1 point) let $f = \frac{1}{5x + 4}$ and find the values below 1. $f(x + h)=$ 2. $(f(x + h)-f(x))=$ 3. $\lim_{h\to0}\frac{f(x + h)-f(x)}{h}=$ 4. find the equation of the line tangent to the graph of $f$ at $x = 1$. $y=$ note: you can earn partial credit on this problem.

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$: $f(x + h)=\frac{1}{5(x + h)+4}=\frac{1}{5x+5h + 4}$

Step2: Find $f(x + h)-f(x)$

\[

$$\begin{align*} f(x + h)-f(x)&=\frac{1}{5x+5h + 4}-\frac{1}{5x + 4}\\ &=\frac{5x + 4-(5x+5h + 4)}{(5x+5h + 4)(5x + 4)}\\ &=\frac{5x + 4-5x-5h - 4}{(5x+5h + 4)(5x + 4)}\\ &=\frac{-5h}{(5x+5h + 4)(5x + 4)} \end{align*}$$

\]

Step3: Find $\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}$
\[

$$\begin{align*} \lim_{h ightarrow0}\frac{f(x + h)-f(x)}{h}&=\lim_{h ightarrow0}\frac{\frac{-5h}{(5x+5h + 4)(5x + 4)}}{h}\\ &=\lim_{h ightarrow0}\frac{-5}{(5x+5h + 4)(5x + 4)}\\ &=-\frac{5}{(5x + 4)^2} \end{align*}$$

\]

Step4: Find the equation of the tangent - line at $x = 1$

First, find $f(1)$: $f(1)=\frac{1}{5\times1+4}=\frac{1}{9}$
Then, find the slope of the tangent - line at $x = 1$. Substitute $x = 1$ into the derivative: $m=-\frac{5}{(5\times1 + 4)^2}=-\frac{5}{81}$
Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(1,\frac{1}{9})$ and $m = -\frac{5}{81}$, we have:
\[

$$\begin{align*} y-\frac{1}{9}&=-\frac{5}{81}(x - 1)\\ y-\frac{1}{9}&=-\frac{5}{81}x+\frac{5}{81}\\ y&=-\frac{5}{81}x+\frac{5}{81}+\frac{9}{81}\\ y&=-\frac{5}{81}x+\frac{14}{81} \end{align*}$$

\]

Answer:

1. $\frac{1}{5x+5h + 4}$
2. $\frac{-5h}{(5x+5h + 4)(5x + 4)}$
3. $-\frac{5}{(5x + 4)^2}$
4. $y=-\frac{5}{81}x+\frac{14}{81}$