Derivatives and its relationship to increasing and decreasing functions

Summary

Suppose that \(f\) is differentiable on an interval \(I\) (open or closed, bounded or unbounded). Then

• \(f'(x)≥0\) for all \(x∈I\) \(⟺\) \(f\) is increasing on \(I\) .
• \(f'(x)≤0\) for all \(x∈I\) \(⟺\) \(f\) is decreasing on \(I\) .
• \(f'(x)>0\) for all \(x∈I\) \(⇒\) \(f\) is strictly increasing on \(I\) (opposite is not necessarily true)
• \(f'(x)<0\) for all \(x∈I\) \(⇒\) \(f\) is strictly decreasing on \(I\) (opposite is not necessarily true)
• Example: If \(f\left( x \right)=e^x\) then \(f'\left( x \right)=e^x>0\) and \(f\) is strictly increasing.
• Example: If \(f\left( x \right)=x^3\) then \(f\left( x \right)\) is strictly increasing but \(f'\left( x \right)>0\) is false since \(f'\left( 0 \right)=0\) .
• Technical note: If \(I\) is a closed interval \(\left[ a,b \right]\) then \(f\) need not be differentiable at the boundary point for these results to hold
• Example: If \(f\left( x \right)=\sqrt{x}\) with domain \(\left[ 0,∞ \right)\) then \(f'\left( x \right)= \frac{1}{2\sqrt{x}}\) and \(f\) is not differentiable at \(x=0\) . However, \(f\) is still strictly increasing on \(\left[ 0,∞ \right)\) .