Question

find the equation of the line tangent to the graph of (f(x)=3csc(x)-2sqrt{3}-\frac{2pi}{3}+1) at (x = \frac{pi}{3}). submit your answer by dragging the red point to the (y)-intercept of the tangent line. provide your answer below:

Explanation:

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

The derivative of \(y = \csc(x)\) is \(y'=-\csc(x)\cot(x)\). So, \(f'(x)=- 3\csc(x)\cot(x)\).

Step2: Evaluate the derivative at \(x = \frac{\pi}{3}\)

We know that \(\csc(\frac{\pi}{3})=\frac{2}{\sqrt{3}}\) and \(\cot(\frac{\pi}{3})=\frac{1}{\sqrt{3}}\). Then \(f'(\frac{\pi}{3})=-3\times\frac{2}{\sqrt{3}}\times\frac{1}{\sqrt{3}}=- 2\).

Step3: Use the point - slope form \(y - y_1=m(x - x_1)\)

The point \((x_1,y_1)\) on the curve is \((\frac{\pi}{3},f(\frac{\pi}{3}))\). \(f(\frac{\pi}{3})=3\csc(\frac{\pi}{3})-2\sqrt{3}-\frac{2\pi}{3}+1=3\times\frac{2}{\sqrt{3}}-2\sqrt{3}-\frac{2\pi}{3}+1 = 2\sqrt{3}-2\sqrt{3}-\frac{2\pi}{3}+1=1-\frac{2\pi}{3}\approx1 - 2.094=-1.094\). The slope \(m = f'(\frac{\pi}{3})=-2\). The point - slope form is \(y-(1 - \frac{2\pi}{3})=-2(x - \frac{\pi}{3})\).

Step4: Rewrite in slope - intercept form \(y=mx + b\)

\[

$$\begin{align*} y-(1-\frac{2\pi}{3})&=-2(x - \frac{\pi}{3})\\ y-(1-\frac{2\pi}{3})&=-2x+\frac{2\pi}{3}\\ y&=-2x + 1 \end{align*}$$

\]
The \(y\) - intercept \(b = 1\).

Answer:

1