1. find the absolute maximum and minimum of the function ( f(x)=2xsqrt{8 - x^{2}} ) on the interval (…

Explanation:

Step1: Differentiate the function

Use the product rule ((uv)^\prime = u^\prime v+uv^\prime), where (u = 2x), (u^\prime=2), (v=\sqrt{8 - x^{2}}=(8 - x^{2})^{\frac{1}{2}}), (v^\prime=\frac{-2x}{2\sqrt{8 - x^{2}}}=\frac{-x}{\sqrt{8 - x^{2}}}).

(f^\prime(x)=2\sqrt{8 - x^{2}}+2x\times\frac{-x}{\sqrt{8 - x^{2}}}=\frac{16 - 2x^{2}-2x^{2}}{\sqrt{8 - x^{2}}}=\frac{16 - 4x^{2}}{\sqrt{8 - x^{2}}})

Step2: Find critical points

Set (f^\prime(x) = 0), then (16-4x^{2}=0) (since the denominator (\sqrt{8 - x^{2}}\neq0) for (x\in(-\sqrt{8},\sqrt{8}))).

(4x^{2}=16), (x^{2} = 4), (x=\pm2)

Step3: Evaluate the function at critical and endpoints

• For (x =-\sqrt{8}): (f(-\sqrt{8})=2\times(-\sqrt{8})\times\sqrt{8-8}=0)
• For (x=- 2): (f(-2)=2\times(-2)\times\sqrt{8 - 4}=-4\times2=-8)
• For (x = 2): (f(2)=2\times2\times\sqrt{8 - 4}=4\times2 = 8)
• For (x=\sqrt{8}): (f(\sqrt{8})=2\times\sqrt{8}\times\sqrt{8 - 8}=0)

Answer:

The absolute maximum value is (8) and the absolute minimum value is (-8)