Proof of convexity of a function

Question

I have to prove that function $J(x)=e^{x^3+x^2+1}$ is convex on $[0,\infty]$.

I used a Theorem which says: **$U\subset R^n$ is convex set with non-empty interior and $J\in C^2(U)$. Function $J$ is convex on $U$ if and only if $\langle J''(u)ξ,ξ\rangle\geq 0$ for all $u\in U$ and all $ξ\in R^n$.

I got that $J''(u)=(6x+2)e^{x^3+x^2+1}+(3x^2+2x)^2e^{x^3+x^2+1}$ which is $\geq 0$ for all $x\ [0,\infty]$.
Hence, $J(u)$ is convex on $[0,\infty]$.

Is this ok?

Answer 1

If $g$ is convex, then $\exp \circ g$ is convex ($\exp(x) = e^x$, and $(\exp \circ g)''(x) = e^{g(x)}((g'(x))^2 + g''(x))$, which is clearly non-negative).

If $g(x) = x^3+x^2 +1$, then $g''(x) = 6 x +2$ which is non-negative on $[0,\infty)$.

It follows that $J= \exp \circ g$ is convex (on $[0,\infty)$).