Răspuns :
[tex]f(x)= e^{ \frac{1}{x} };f'(x)=- \frac{1}{ x^{2} } e^{ \frac{1}{x} } ;f''(x)= \frac{2}{x^3} e^{ \frac{1}{x} }+ \frac{1}{x^4} e^{ \frac{1}{x} } \\ f''(x)=0 \\ \frac{1}{x^3} e^{ \frac{1}{x} }(2+ \frac{1}{x})=0 \\ 2+ \frac{1}{x}=0 \\ \frac{2x+1}{x}=0;2x+1=0;x=- \frac{1}{2} [/tex]