[tex]f(x)=2x^2-5x+1 \\ f'(x)=4x-5 \\ f'(x)=0\Leftrightarrow 4x-5=0\Leftrightarrow x= \frac{5}{4} \\ f( \frac{5}{4})=2\cdot \frac{25}{16}-5\cdot \frac{5}{4}+1= \frac{25}{8}+ \frac{50}{8}+ \frac{8}{8}= \frac{83}{8} \\ Punctl\ de\ extrem\ are\ coordonatele\ ( \frac{5}{4}; \frac{83}{8}) [/tex]
