Is this sufficient to show that $f(p) \leq g(p)$ holds for all $p \geq 0$?

Question

$f(p) = 1 - p, g(p) = e^{-p}$ I have shown that $f(0) = g(0)$ and $f'(p) \leq g'(p)$. Is this sufficient to show that $f(p) \leq g(p)$ holds for all $p \geq 0$? Is there a specific theorem that I need to state?

Thanks!

Answer 1

Yes, this is a result from the fundamental theorem of calculus, combined with monotonity of integrals. Basically if $f,g$ are continuously differentiable so that $f(x_0)\leq g(x_0)$ and $f'(x)\leq g'(x)$ for $x>x_0$.

Then $$ f(x) = f(x_0)+\int_{x_0}^x f'\,\mathrm d x \leq g(x_0)+\int_{x_0}^x g'\,\mathrm d x = g(x) $$

But in your example it suffices to see that you only need to consider $p<1$ (as $g>0$). Then $f(p)\leq 1$ and $g(p)\geq 1$.