이계도함수 판정법을 이용하여 다음 함수의 극점을 판별하여라.
$$f(x)=x\sqrt{8-x^2}$$
풀이
\(\displaystyle f(x)=x(8-x^2)^\frac{1}{2}\)
\(\displaystyle f'(x)=x'(8-x^2)^\frac{1}{2}+x\left((8-x^2)^\frac{1}{2}\right)’\)
\(\displaystyle=(8-x^2)^\frac{1}{2}+x\left(\frac{1}{2}(8-x^2)^{-\frac{1}{2}}(-2x)\right)\)
\(\displaystyle=(8-x^2)^\frac{1}{2}-x^2(8-x^2)^{-\frac{1}{2}}\)
\(\displaystyle=\sqrt{8-x^2}-\frac{x^2}{\sqrt{8-x^2}}\)
\(\displaystyle=\frac{8-x^2-x^2}{\sqrt{8-x^2}}=\frac{8-2x^2}{\sqrt{8-x^2}}\)
\(\displaystyle=\frac{-2(x^2-4)}{\sqrt{8-x^2}}=\frac{-2(x+2)(x-2)}{\sqrt{8-x^2}}\)
\(\displaystyle f'{}'(x)=\frac{(8-2x^2)’\sqrt{8-x^2}-(8-2x^2)(\sqrt{8-x^2})’}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}-(8-2x^2)((8-x^2)^\frac{1}{2})’}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}-(8-2x^2)(\frac{1}{2}(8-x^2)^{-\frac{1}{2}}(-2x))}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}+x(8-2x^2)((8-x^2)^{-\frac{1}{2}})}{8-x^2}\)
\(\displaystyle=\frac{-4x\sqrt{8-x^2}+\frac{x(8-2x^2)}{\sqrt{8-x^2}}}{8-x^2}\)
\(\displaystyle=\frac{\frac{-4x(8-x^2)+x(8-2x^2)}{\sqrt{8-x^2}}}{8-x^2}\)
\(\displaystyle=\frac{-4x(8-x^2)+x(8-2x^2)}{\sqrt{8-x^2}(8-x^2)}\)
\(\displaystyle=\frac{4x^3-32x-2x^3+8x}{\sqrt{8-x^2}(8-x^2)}\)
\(\displaystyle=\frac{2x^3-24x}{\sqrt{8-x^2}(8-x^2)}\)
\(\displaystyle f'(x)=0\)인 \(x\)의 값은 \(-2\), \(2\)이다.
\(\displaystyle f'{}'(-2)=\frac{2(-2)^3-24(-2)}{\sqrt{8-(-2)^2}(8-(-2)^2)}\)
\(\displaystyle=\frac{2(-8)+48}{\sqrt{8-4}(8-4)}=\frac{-16+48}{\sqrt{4}(4)}\)
\(\displaystyle=\frac{32}{2(4)}=\frac{32}{8}=4>0\)
\(\displaystyle f'{}'(2)=\frac{2(2)^3-24(2)}{\sqrt{8-(2)^2}(8-(2)^2)}\)
\(\displaystyle=\frac{2(8)-48}{\sqrt{8-4}(8-4)}\)
\(\displaystyle=\frac{16-48}{\sqrt{4}(4)}=\frac{-32}{2(4)}\)
\(\displaystyle=\frac{-32}{8}=-4<0\)
\(f(-2)=(-2)\sqrt{8-(-2)^2}=-2\sqrt{8-4}=-2\sqrt{4}=-2(2)=-4\)
\(f(2)=(2)\sqrt{8-(2)^2}=2\sqrt{8-4}=2\sqrt{4}=2(2)=4\)
따라서 \(f(x)\)는 \(x=-2\)일 때 극소값 \(-4\), \(x=2\)일 때 극대값 \(4\)를 갖는다.