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 $\left[ 0,∞ \right)$ .