1. find the absolute maximum and minimum for these closed intervals:\na. $f(x)=(\\sin^{-1}x)-2x$ $-1,1$

Explanation:

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

The derivative of (y = \sin^{-1}x) is (y^\prime=\frac{1}{\sqrt{1 - x^{2}}}), and the derivative of (y=-2x) is (y^\prime=-2). Using the sum - rule ((u + v)^\prime=u^\prime + v^\prime) (where (u=\sin^{-1}x) and (v = - 2x)), we get (f^\prime(x)=\frac{1}{\sqrt{1 - x^{2}}}-2).

Step2: Find the critical points

Set (f^\prime(x) = 0), so (\frac{1}{\sqrt{1 - x^{2}}}-2=0). Rearrange the equation: (\frac{1}{\sqrt{1 - x^{2}}}=2), then (\sqrt{1 - x^{2}}=\frac{1}{2}), and (1 - x^{2}=\frac{1}{4}), (x^{2}=\frac{3}{4}), (x=\pm\frac{\sqrt{3}}{2}). Both (x = \frac{\sqrt{3}}{2}) and (x=-\frac{\sqrt{3}}{2}) are in the interval ([-1,1]).

Step3: Evaluate (f(x)) at the critical points and endpoints

• When (x=-1): (f(-1)=\sin^{-1}(-1)-2\times(-1)=-\frac{\pi}{2}+2\approx2 - 1.57=0.43).
• When (x =-\frac{\sqrt{3}}{2}): (f\left(-\frac{\sqrt{3}}{2}\right)=\sin^{-1}\left(-\frac{\sqrt{3}}{2}\right)-2\times\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}+\sqrt{3}\approx - 1.05+1.73 = 0.68).
• When (x=\frac{\sqrt{3}}{2}): (f\left(\frac{\sqrt{3}}{2}\right)=\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)-2\times\frac{\sqrt{3}}{2}=\frac{\pi}{3}-\sqrt{3}\approx1.05 - 1.73=-0.68).
• When (x = 1): (f(1)=\sin^{-1}(1)-2\times1=\frac{\pi}{2}-2\approx1.57 - 2=-0.43).

Answer:

The absolute maximum value of (f(x)) on the interval ([-1,1]) is (f\left(-\frac{\sqrt{3}}{2}\right)=-\frac{\pi}{3}+\sqrt{3}\approx0.68), and the absolute minimum value is (f\left(\frac{\sqrt{3}}{2}\right)=\frac{\pi}{3}-\sqrt{3}\approx - 0.68).