Derivatives and its relationship to convex and concave functions

Summary

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

• \(f''(x)≥0\) for all \(x∈I\) \(⟺\) \(f\) is convex in \(I\)
• \(f''(x)≤0\) for all \(x∈I\) \(⟺f\) is concave in \(I\)
• If \(f''(x)>0\) for all \(x∈I\) \(⇒\) \(f\) is strictly convex in \(I\) (opposite is not necessarily true)
• If \(f''(x)<0\) for all \(x∈I\) \(⇒\) \(f\) is strictly concave in \(I\) (opposite is not necessarily true)
• Examples:
• \(f(x)=x^2\) , \(f''(x)=2\) and the function is strictly convex on any interval.
• \(f(x)=-x^2\) , \(f''(x)=-2\) and the function is strictly concave on any interval.
• \(f(x)=x\) , \(f''(x)=0\) and the function is concave and convex on any interval.
• \(f(x)=x^4\) is strictly convex on any interval but \(f''(x)>0\) is false. \(f''(x)=12x^2\) and \(f^{''}\left( 0 \right)=0\) .
• 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, \(f^{''}\left( x \right)=-x^{-3/2}/4<0\) for \(x>0\) and \(f\) is still strictly concave on \(\left[ 0,∞ \right)\) .