Respostas

2013-12-17T17:16:54-02:00
f(x)=\frac{x^2}{x-3}

Candidatos a pontos críticos

f'(x)=\frac{2x(x-3)-2x}{(x-3)^2}=\frac{2x^2-8x}{(x-3)^2}=0\newline 2x^2-8x=x(2x-8)=0\newline x=0\newline 2x-8=0.:\ x= \frac{8}{2}=4

Teste da segunda derivada
f''(x)=\frac{2x^2-8x}{(x-3)^2}=4x-8-(2x^2-8x).2.(x-3).1\newline f''(0)=4.0-8-(2.0^2-8.0).2.(0-3).1=-8<0\ ponto\ de\ m\´aximo

f''(4)=4.4-8-(2.4^2-8.4).2.(4-3).1=\newline 16-8-(32-32).2.1.1=8>0\ ponto\ de\ m\´inimo

Soma dos quadrados

f(0)=\frac{0^2}{0-3}=0\newline f(4)=\frac{4^2}{4-3}=\frac{16}{1}=16\newline 0^2+16^2=256

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