Using the mean value theorem to prove that $f(x) \leq e^{-\frac{x}{2}}$

Question

Here is the link to the question I was trying to prove:

Why $f(x) = 0$ is not excluded from the codomain?

And here is its statement:

Suppose $f:[0, \infty) \to \mathbb R$ is a continuously differentiable function and

$\bullet f(0) =1$.

$\bullet f'(x) \leq \frac{-1}{2} f(x)$ for all $x \geq 0.$

Prove that $f(x) \leq e^{\frac{-x}{2}}$ for $x\geq 0.$

My question is:

I got a hint (in the link above from Andrew) to consider the function $g(x)$ and specifically, this was the hint:

"Instead, it's cleaner to introduce the function $g(x) = e^{x/2}f(x)$, and use the product rule and $f'(x) \leq -\frac{1}{2}f(x)$ to show $g'(x) \leq 0$ for all non-negative $x$. The mean value theorem together with $g(0) = 1$ then guarantees $g(x) \leq 1$ for all non-negative $x$. "

I managed to show that $g'(0) \leq 0$ for all non-negative $x$. But I do not know how "The mean value theorem together with $g(0) = 1$ then guarantees $g(x) \leq 1$ for all non-negative $x$." Could someone show me the details of this proof please?

Answer 1

Let $x\geq 0$, by the MVT there exists a $c\in (0,x)$ such that $$ \frac{g(x)-g(0)}{x} = g'(c) \leq 0 \Rightarrow g(x)\leq g(0) = 1$$

Answer 2

So $g(x) = e^{x/2}f(x)$ thus $$g'(x) = \frac{1}2 e^{x/2}f(x) + e^{x/2}f'(x)\leq \frac{1}2 e^{x/2}f(x) + (-1/2)e^{x/2}f(x)=0 $$

Now for any $[a,b]\subset [0,\infty)$ we have from MVT for some $x\in (a,b)$ $$\frac{g(b)-g(a)}{b-a} = g'(c)\leq 0$$ If $a=0$ we get $$\frac{g(b)-1}{b} \leq 0$$ so $$g(b) \leq 1$$ That for any $b>0$ hence $f(x)\leq e^{-x/2}$ for any $x>0$