Question

Sa se tetermine coordonatele punctului de extrem al functiei f:R>R. f(x)=2x la a2a -5x+1

Answer 1

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