Question

Sa se rezolve in R ecuatia f'(x)=0,unde f:D-->R:
a) f(x)=5x^2-3x+8;

Answer 1

$f:D\rightarrow R \\ f(x)=5x^2-3x+8 \\ f'(x)=10x-3 \\ f'(x)=0\leftrightarrow 10x-3=0\rightarrow x= \frac{3}{10}=0.3$