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

Explanation:

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

The derivative of (y=\sin^{-1}(x)) is (y'=\frac{1}{\sqrt{1 - x^{2}}}) (using the formula for the derivative of the inverse - sine function ((\sin^{-1}(u))'=\frac{u'}{\sqrt{1 - u^{2}}}), here (u = x) and (u'=1)).

Step2: Evaluate the derivative at (x =-\frac{1}{2})

Substitute (x =-\frac{1}{2}) into (y'=\frac{1}{\sqrt{1 - x^{2}}}). Then (y'\mid_{x =-\frac{1}{2}}=\frac{1}{\sqrt{1-\left(-\frac{1}{2}\right)^{2}}}=\frac{1}{\sqrt{1-\frac{1}{4}}}=\frac{1}{\sqrt{\frac{3}{4}}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}). So the slope (m) of the tangent line is (\frac{2\sqrt{3}}{3}).

Step3: Find the (y) - coordinate when (x =-\frac{1}{2})

If (x =-\frac{1}{2}), then (y=\sin^{-1}\left(-\frac{1}{2}\right)=-\frac{\pi}{6}) (since (\sin\left(-\frac{\pi}{6}\right)=-\frac{1}{2}) and the range of (y = \sin^{-1}(x)) is (\left[-\frac{\pi}{2},\frac{\pi}{2}\right])).

Step4: Use the point - slope form (y - y_{1}=m(x - x_{1}))

Here (x_{1}=-\frac{1}{2}), (y_{1}=-\frac{\pi}{6}), and (m = \frac{2\sqrt{3}}{3}). [ \begin{align*} y-\left(-\frac{\pi}{6}\right)&=\frac{2\sqrt{3}}{3}\left(x+\frac{1}{2}\right)\ y+\frac{\pi}{6}&=\frac{2\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}\ y&=\frac{2\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}-\frac{\pi}{6} \end{align*} ]

Answer:

(y=\frac{2\sqrt{3}}{3}x+\frac{\sqrt{3}}{3}-\frac{\pi}{6})