# First derivative test

We have that Let $f$ be continuous on an interval $I$ and differentiable on the interior of $I$. If $f'(x)>0$ for all $x\in I$, then $f$ is increasing on $I$. If $f'(x)<0$ for all $x\in I$, then $f$ is decreasing on $I$. If $f'(x)=0$ for all $x\in I$, then $f$ is constant on $I$.

My question is can we conclude in opposite direction as well, i.e., If $f$ is increasing on $I$, then $f'(x)>0$ for all $x\in I$, in all three cases? If not why?

• – Cameron Buie Nov 6 '12 at 6:43

In the third case, it will work, a constant function on $I$ has derivative $0$. But if $f\colon I \to \mathbb R$ is increasing, you will in general only have $f' \ge 0$. Consider for example $f\colon\mathbb R \to \mathbb R$ given by $f(x) = x^3$, it's derivative $f'(x) = 3x^2$ fulfills $f'(0) = 0$.
We cannot in the first two cases. Consider $f(x)=x^3$ and $g(x)=-x^3$, on the interval $I=[-1,1]$, to see that increasing doesn't imply positive derivative, and that decreasing doesn't imply negative derivative.
However, if $f$ is constant, we can conclude that the derivative is $0$--this follows readily from the difference quotient definition of derivative.