If $f:[0,\infty) \to \mathbb{R}$ be a differentiable function with $f(0)=1$ . . . Show that $f'(x)\geq e^x$ for all $x>0$. If $f:[0,\infty) \to \mathbb{R}$ be a differentiable function with $f(0)=1$ and $f'(x)\geq f(x)$ for all $x>0$. Show that $f'(x)\geq e^x$ for all $x>0$.
Hint: Consider $g(x)=e^{-x}f(x)$.
My main concern is what does $g(x)$ have to do with anything. I feel if I understood its relevance, then I could solve the problem. I'm just not sure how I can declare it's relevance to the problem at hand.
Edit:
charlestoncrabb is correct on the proof, I'm just not certain of the relevance of $g(x)$
 A: Here is an answer for why it must be that $f(x)\geq e^x$:
This is in some sense a "reverse Grönwall inequality". Compute $g'(x)$, as suggested, using the product rule:
$$g'(x)=e^{-x}(f'(x)-f(x))\geq 0.$$
Since $f(0)=1$, we have $g(0)=1$ as well, thus $g$ is a non-decreasing function with $g(0)=1$, implying that $g(x)\geq1$, so that $e^{-x}f(x)\geq1\Longrightarrow f(x)\geq e^x$. Thus $f'(x)\geq e^x$ as well.
(Edit: Remember that we assumed $f'(x)\geq f(x)$ to begin with, so $f(x)\geq e^x\Longrightarrow f'(x)\geq e^x$)
(2nd Edit: Also notice the $f(0)=1$ assumption is essential, as throwing it out yields the counterexample $f\equiv0$)
(3rd Edit: The relevance of $g$ is that we have shown $g(x)\geq 1$ for all $x$, which is equivalent the desired inequality)
A: Here is another answer.
Suppose $A:=\{x \geq 0:f(x) \leq0 \}$ is non - empty set and take $c := \inf A$
Obviously, $f(c) \leq 0$ (Since $f$ is conti)
By Mean value theorem, we can take $a \in (0,c)$ such that $f'(a) \leq 0$.
However, since $f'(x) \geq f(x)$, $f'(a) \geq f(a)$ so $a \in A$.
This is contradiction to minimality of $c$.
So, $A$ is empty and this implies that $f$ is positive function.
Now, let's get this over with.
$$f'(x) \geq f(x) \Leftrightarrow \frac{f'(x)}{f(x)} \geq 1$$
$$\Leftrightarrow \int_0^x \frac{f'(t)}{f(t)} dt \geq x \Leftrightarrow \ln f(x) \geq x \Leftrightarrow f(x) \geq e^x $$
Since $f'(x) \geq f(x)$, so we're done. ($f'(x) \geq f(x) \geq e^x , x \geq 0)$
