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ekstremum i punkty przegięcia

: 27 lip 2007, o 00:26
autor: Hania_87
\(\displaystyle{ f(x)=3x+\frac{1}{x}}\)

ekstremum i punkty przegięcia

: 27 lip 2007, o 01:58
autor: soku11
\(\displaystyle{ f(x)=3x+\frac{1}{x}\ \ D_f=R\backslash\{0\} \\
f'(x)=3-\frac{1}{x^{2}}\ \ D_{f'}=R\backslash\{0\} \\
f''(x)=\frac{2}{x^{3}}\ \ D_{f''}=R\backslash\{0\} \\
f'(x)=0\ \iff\ 3-\frac{1}{x^{2}}=0\\
\frac{3x^{2}-1}{x^{2}}=0\\
3x^{2}-1=0\\
3(x^{2}-\frac{1}{3})=0\\
3(x-\frac{\sqrt{3}}{3})(x+\frac{\sqrt{3}}{3})=0\\
f_{min}=f(\frac{\sqrt{3}}{3})\\
f_{max}=f(-\frac{\sqrt{3}}{3})\\}\)


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