You can simply show that f is strictly increasing over )0,+oo( and that its limit in 0 is -oo, and its limit in +oo is +oo
I need to prove that f(x)=e^x * ln x gets every value in R exactly once. I managed to prove that f(x) gets every value in R , i.e. , for each y in R there is x in (0,inf) such as f(x)=y . But when I tried to show that f gets every value in R exactly once , i.e. showing that f is bijective , I kinda got stuck...
I tried to assume that there are x_1 , x_2 in (0,inf) such as f(x_1)=f(x_2).
f is differentiable in (0,inf) as a product of differentiable functions in (0,inf) , and therefore , from Rolle's theorem I can say that there is a point c in (x_1 , x_2) such as f'(c)=0 , i.e. , (e^x / x) + e^x*ln x = 0.
if x >= 1, then it is a contradiction , becuase for x>=1 , (e^x / x) + e^x*ln x is always positive. But if 0<x<1 , then e^x*ln x is negetive , and e^x / x is positive of course , so I can't say that f's derivative is always different from zero.. I tried to prove it without Rolle's theorem as well, but I didn't managed to... Any clues ?
I know. f's derivative is always positive at the interval (1,inf), and therefore f is strictly increasing there.
But I don't know nothing about f's derivative at the interval (0,1) , becuase for each 0<x<1 , e^x /x is positive , and e^x*ln x is negative , so I don't really know much about f's derivative at the interval (0,1) , that's why I am stuck...