Question

question
find the equation of the line tangent to the graph of

$f(x)=-3 - 3csc(x)$ at $x =-\frac{pi}{3}$

(enter an exact answer in $y = mx + b$ form.)

provide your answer below:

$y=square$

Explanation:

Step1: Find the derivative of $f(x)$

The derivative of $\csc(x)$ is $-\csc(x)\cot(x)$. So, $f^\prime(x)=3\csc(x)\cot(x)$.

Step2: Evaluate $f(x)$ and $f^\prime(x)$ at $x =-\frac{\pi}{3}$

First, find $f(-\frac{\pi}{3})$:
\[

$$\begin{align*} f(-\frac{\pi}{3})&=-3 - 3\csc(-\frac{\pi}{3})\\ &=-3+3\csc(\frac{\pi}{3})\\ &=-3 + 3\times\frac{2}{\sqrt{3}}\\ &=-3 + 2\sqrt{3} \end{align*}$$

\]
Next, find $f^\prime(-\frac{\pi}{3})$:
\[

$$\begin{align*} f^\prime(-\frac{\pi}{3})&=3\csc(-\frac{\pi}{3})\cot(-\frac{\pi}{3})\\ &=- 3\csc(\frac{\pi}{3})\cot(\frac{\pi}{3})\\ &=-3\times\frac{2}{\sqrt{3}}\times\frac{1}{\sqrt{3}}\\ &=- 2 \end{align*}$$

\]

Step3: Use the point - slope form $y - y_1=m(x - x_1)$ to find the equation of the tangent line

Here, $m = f^\prime(-\frac{\pi}{3})=-2$, $x_1 =-\frac{\pi}{3}$, and $y_1=f(-\frac{\pi}{3})=-3 + 2\sqrt{3}$.
\[

$$\begin{align*} y-(-3 + 2\sqrt{3})&=-2(x+\frac{\pi}{3})\\ y+3 - 2\sqrt{3}&=-2x-\frac{2\pi}{3}\\ y&=-2x-\frac{2\pi}{3}-3 + 2\sqrt{3} \end{align*}$$

\]

Answer:

$y=-2x-\frac{2\pi}{3}-3 + 2\sqrt{3}$