4. for each of the following functions: a) find an equation of the tangent line to the curve at the given…

Explanation:

Step1: Recall derivative formulas

The derivative of $y = \arctan x$ is $y'=\frac{1}{1 + x^{2}}$, and the derivative of $y=\sin^{-1}x$ is $y'=\frac{1}{\sqrt{1 - x^{2}}}$.

Step2 - i - a: Find the slope of the tangent line for $s(t)=\arctan x$ at $x = - 1$

First, find the derivative $s'(x)=\frac{1}{1 + x^{2}}$. When $x=-1$, $s'(-1)=\frac{1}{1+(-1)^{2}}=\frac{1}{2}$. Also, $s(-1)=\arctan(-1)=-\frac{\pi}{4}$. The point - slope form of a line is $y - y_0=m(x - x_0)$. Here, $x_0=-1,y_0 =-\frac{\pi}{4},m=\frac{1}{2}$. So the equation of the tangent line is $y+\frac{\pi}{4}=\frac{1}{2}(x + 1)$, which simplifies to $y=\frac{1}{2}x+\frac{1}{2}-\frac{\pi}{4}$.

Step2 - i - b: Find when the tangent line is horizontal for $s(t)=\arctan x$

Set $s'(x)=\frac{1}{1 + x^{2}} = 0$. Since the numerator is 1, there are no real - valued solutions for $x$ because $1+x^{2}\geq1$ for all real $x$, so there are no values of $x$ for which the tangent line is horizontal.

Step3 - ii - a: Find the slope of the tangent line for $f(x)=\sin^{-1}x$ at $x=\frac{1}{2}$

The derivative $f'(x)=\frac{1}{\sqrt{1 - x^{2}}}$. When $x = \frac{1}{2}$, $f'(\frac{1}{2})=\frac{1}{\sqrt{1-(\frac{1}{2})^{2}}}=\frac{1}{\sqrt{\frac{3}{4}}}=\frac{2}{\sqrt{3}}=\frac{2\sqrt{3}}{3}$. And $f(\frac{1}{2})=\sin^{-1}(\frac{1}{2})=\frac{\pi}{6}$. Using the point - slope form $y - y_0=m(x - x_0)$ with $x_0=\frac{1}{2},y_0=\frac{\pi}{6},m = \frac{2\sqrt{3}}{3}$, we get $y-\frac{\pi}{6}=\frac{2\sqrt{3}}{3}(x-\frac{1}{2})$, which simplifies to $y=\frac{2\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}+\frac{\pi}{6}$.

Step3 - ii - b: Find when the tangent line is horizontal for $f(x)=\sin^{-1}x$

Set $f'(x)=\frac{1}{\sqrt{1 - x^{2}}}=0$. Since the numerator is 1, there are no real - valued solutions for $x$ because $\sqrt{1 - x^{2}}>0$ for $|x|<1$ and is undefined for $|x|>1$.

Answer:

i - a: $y=\frac{1}{2}x+\frac{1}{2}-\frac{\pi}{4}$ i - b: No real - valued $x$ ii - a: $y=\frac{2\sqrt{3}}{3}x-\frac{\sqrt{3}}{3}+\frac{\pi}{6}$ ii - b: No real - valued $x$