This question was in my test and I am not sure what to do with it
let $f:(0,\infty)\rightarrow \mathbb{R}$ be given by $$f(x)=\log x-x+2$$ then its number of roots of $f$are.
So putting it equal to zero might won't help gives this $x=e^(x-2)$ I am not getting it how to solve it, then I thought of doing its derivatives and find out where it is zero (it is at 1 easy to find) then I tried finding an interval around it so that I will get negative value on one side and positive value on other but I find it usually non-negative (by hit and trial) so what else I can do to solve it.
$f'(x) = 1/x - 1.$
For x:(0,1) $f'(x) >0.$
Now $\lim_{x\to 0^+}f(x)=-\infty$.
And f(x) is monotonically increasing upto x=1 where f(1)=1. So it becomes 0 somewhere between 0 and 1. So here we get our 1st root.
For x>1 :$ f'(x) <0.$
Now the function is monotonically decreasing and$ f(1)=1 $and f(a large number say 100) = - ve. So it again becomes 0 once. So we get our 2nd root. Now it just gets more negative since it goes on decreasing so there's no other root.
So we can conclude that there are 2 roots of the equation.