Question

assignment 4: problem 1 (1 point) suppose $f(x)=\frac{3}{x - 3}$, $f(4)=square$. use this to find the equation of the tangent line to the curve $y=\frac{3}{x - 3}$ at the point $(4,3)$. write you answer in the form $y=mx + b$, where $m$ is the slope and $b$ is the $y$-intercept. the equation of the tangent line is $square$

Explanation:

Step1: Differentiate the function

Use the quotient - rule. If $f(x)=\frac{u}{v}$ where $u = 3$ and $v=x - 3$, then $f'(x)=\frac{u'v - uv'}{v^{2}}$. Since $u'=0$ and $v' = 1$, we have $f'(x)=\frac{0\times(x - 3)-3\times1}{(x - 3)^{2}}=-\frac{3}{(x - 3)^{2}}$.

Step2: Evaluate $f'(4)$

Substitute $x = 4$ into $f'(x)$. So $f'(4)=-\frac{3}{(4 - 3)^{2}}=- 3$.

Step3: Find the equation of the tangent line

The equation of a line is $y - y_1=m(x - x_1)$, where $(x_1,y_1)=(4,3)$ and $m=-3$.
$y - 3=-3(x - 4)$
Expand the right - hand side: $y - 3=-3x+12$.
Add 3 to both sides to get the form $y=mx + b$: $y=-3x + 15$.

Answer:

$f'(4)=-3$
The equation of the tangent line is $y=-3x + 15$