$g(0)=1,g'(x)\geq g(x)$ for all $x>0$ and $g'(x)\leq g(x)$ for all $x<0$. Prove that $g(x)\geq exp(x)$ for all $x\in\mathbb{R}$.

Question

Let $g:\mathbb{R}\rightarrow\mathbb{R}$ be a differentiable function, $g(0)=1,g'(x)\geq g(x)$ for all $x>0$ and $g'(x)\leq g(x)$ for all $x<0$. Prove that $g(x)\geq exp(x)$ for all $x\in\mathbb{R}$.

Could anyone please help me?

Answer 1

Define $f(x)=e^{-x}g(x)$. Then $f(0)=1$, and for all $x> 0$,$$f'(x)=e^{-x}(g'(x)-g(x))\ge 0$$Hence $f(x)\ge1$ for all $x>0$, i.e. $g(x)\ge e^x$ for all $x>0$.

A similar argument shows that $g(x)\ge e^x$ for all $x<0$.