find the equation of the line tangent to the graph of ( f ) at the indicated ( x ) value.\n( f(x)=cos…

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})