Question

homework4: problem 7 (1 point) let $f = \frac{1}{3x + 1}$ 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. preview my answers submit answers you have attempted this problem 0 times. you have unlimited attempts remaining. email instructor page generated october 5, 2025, 11:17:54 pm cdt

Explanation:

Step1: Find $f(x + h)$

Substitute $x+h$ into $f(x)$.
$f(x + h)=\frac{1}{3(x + h)+1}=\frac{1}{3x+3h + 1}$

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

$f(x + h)-f(x)=\frac{1}{3x + 3h+1}-\frac{1}{3x + 1}=\frac{(3x + 1)-(3x + 3h + 1)}{(3x + 1)(3x+3h + 1)}=\frac{3x + 1-3x-3h - 1}{(3x + 1)(3x+3h + 1)}=\frac{-3h}{(3x + 1)(3x+3h + 1)}$

Step3: Find $\lim_{h

ightarrow0}\frac{f(x + h)-f(x)}{h}$
$\lim_{h
ightarrow0}\frac{f(x + h)-f(x)}{h}=\lim_{h
ightarrow0}\frac{\frac{-3h}{(3x + 1)(3x+3h + 1)}}{h}=\lim_{h
ightarrow0}\frac{-3}{(3x + 1)(3x+3h + 1)}$. As $h
ightarrow0$, we get $\frac{-3}{(3x + 1)^2}$

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

First, find $f(1)=\frac{1}{3\times1+1}=\frac{1}{4}$.
Second, find the slope of the tangent - line at $x = 1$. Substitute $x = 1$ into the derivative: $m=\frac{-3}{(3\times1 + 1)^2}=\frac{-3}{16}$.
Using the point - slope form $y - y_1=m(x - x_1)$ with $(x_1,y_1)=(1,\frac{1}{4})$ and $m =-\frac{3}{16}$, we have $y-\frac{1}{4}=-\frac{3}{16}(x - 1)$.
$y-\frac{1}{4}=-\frac{3}{16}x+\frac{3}{16}$.
$y=-\frac{3}{16}x+\frac{3}{16}+\frac{1}{4}=-\frac{3}{16}x+\frac{3 + 4}{16}=-\frac{3}{16}x+\frac{7}{16}$

Answer:

1. $\frac{1}{3x+3h + 1}$
2. $\frac{-3h}{(3x + 1)(3x+3h + 1)}$
3. $\frac{-3}{(3x + 1)^2}$
4. $y=-\frac{3}{16}x+\frac{7}{16}$