Question

find the equation of the line tangent to the graph of ( f ) at the indicated ( x ) value.
( f(x)=cos ^{-1}(2 x) ) at ( x=\frac{1}{4} )

Explanation:

Step1: Find the derivative of \(y = f(x)=\cos^{-1}(2x)\)

Using the formula \(\frac{d}{dx}\cos^{-1}(u)=-\frac{1}{\sqrt{1 - u^{2}}}\cdot\frac{du}{dx}\), where \(u = 2x\) and \(\frac{du}{dx}=2\).
So \(y^\prime=-\frac{2}{\sqrt{1-(2x)^{2}}}\)

Step2: Evaluate the derivative at \(x = \frac{1}{4}\)

Substitute \(x=\frac{1}{4}\) into \(y^\prime\):
\(y^\prime|_{x = \frac{1}{4}}=-\frac{2}{\sqrt{1-(2\times\frac{1}{4})^{2}}}=-\frac{2}{\sqrt{1-\frac{1}{4}}}=-\frac{2}{\sqrt{\frac{3}{4}}}=-\frac{4}{\sqrt{3}}=-\frac{4\sqrt{3}}{3}\)

Step3: Find the \(y\) - value at \(x=\frac{1}{4}\)

\(y = f(\frac{1}{4})=\cos^{-1}(2\times\frac{1}{4})=\cos^{-1}(\frac{1}{2})=\frac{\pi}{3}\)

Step4: Use the point - slope form \(y - y_1=m(x - x_1)\)

Here \(x_1=\frac{1}{4}\), \(y_1=\frac{\pi}{3}\) and \(m =-\frac{4\sqrt{3}}{3}\)
\(y-\frac{\pi}{3}=-\frac{4\sqrt{3}}{3}(x - \frac{1}{4})\)
\(y=-\frac{4\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}+\frac{\pi}{3}\)

Answer:

\(y =-\frac{4\sqrt{3}}{3}x+\frac{\sqrt{3}+\pi}{3}\)