Question

Sa se rezolve in R ecuatia f'(x)=0 unde f:D-->R:
f(x)=lg(x^2-x);

Answer 1

$f:D\rightarrow R \\ f(x)=\lg (x^2-x) \\ f'(x)= \frac{2x-1}{(x^2-x)\ln 10} \\ f'(x)=0\leftrightarrow \frac{2x-1}{(x^2-x)\ln 10}=0\rightarrow 2x-1=0\rightarrow x= \frac{1}{2}=0.5$