Example 1
Consider the function $y(x)=x^3+2x$ on the interval $x\geq 0$. It holds that
- $y'(x)=3x^2+2> 0$ for every $x\geq 0$;
- $y''(x)=6x\geq 0$ for every $x\geq 0$.
From the second-order condition it follows that $y(x)$ is convex on the interval $x\geq 0$.
Example 2
Consider the function $y(x)=-2+\sqrt{x+1}$. From $y(x)=-2+(x+1)^{\frac{1}{2}}$ it follows that
- $y'(x)=\frac{1}{2}(x+1)^{-\frac{1}{2}} =\frac{1}{2\sqrt{x+1}}> 0$ for every $x> -1$;
- $y''(x)=-\frac{1}{4}(x+1)^{-\frac{3}{2}}=-\frac{1}{4(x+1)\sqrt{x+1}} < 0$ for every $x> -1$.
From the second-order condition it follows that $y(x)$ is concave on the interval $x>-1$.