determine the location and value of the absolute extreme values of ( f ) on the given interval, if they…

Explanation:

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

The derivative of (y = 3x^{2}+3\sin^{- 1}x) is (f^{\prime}(x)=6x+\frac{3}{\sqrt{1 - x^{2}}}), where (x\in(-1,1)).

Step2: Find the critical points

Set (f^{\prime}(x) = 0), so (6x+\frac{3}{\sqrt{1 - x^{2}}}=0). Multiply through by (\sqrt{1 - x^{2}}) (since (\sqrt{1 - x^{2}}>0) for (x\in(-1,1))) to get (6x\sqrt{1 - x^{2}}+3 = 0). Let (t=\sqrt{1 - x^{2}}), then (x=\pm\sqrt{1 - t^{2}}). The equation becomes (6x t+3 = 0), or (x t=-\frac{1}{2}). Substituting back (t = \sqrt{1 - x^{2}}), we have (x\sqrt{1 - x^{2}}=-\frac{1}{2}). Square both sides: (x^{2}(1 - x^{2})=\frac{1}{4}), let (u = x^{2}), then (u - u^{2}=\frac{1}{4}), or (u^{2}-u+\frac{1}{4}=0). Using the quadratic formula (u=\frac{1\pm\sqrt{1 - 1}}{2}=\frac{1}{2}). Since (x\sqrt{1 - x^{2}}<0), (x=-\frac{\sqrt{2}}{2}).

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

• At (x=-1): (f(-1)=3\times(-1)^{2}+3\sin^{-1}(-1)=3 - \frac{3\pi}{2}).
• At (x =-\frac{\sqrt{2}}{2}): (f\left(-\frac{\sqrt{2}}{2}\right)=3\times\frac{1}{2}+3\sin^{-1}\left(-\frac{\sqrt{2}}{2}\right)=\frac{3}{2}-\frac{3\pi}{4}).
• At (x = 1): (f(1)=3\times1^{2}+3\sin^{-1}(1)=3+\frac{3\pi}{2}).

Answer:

C. The absolute maximum value is (3+\frac{3\pi}{2}) at (x = 1) and the absolute minimum value is (3-\frac{3\pi}{2}) at (x=-1)