Pretend that you don’t know about the function $ e^x$ (so pretend to erase it and everything you know about it from your memory, for argument’s sake).
You want to draw graphs of functions that satisfy $f’(x) = f(x)$. Not precisely, but you want to know it’s qualitative properties.
$f(x) = 0 \forall x$ is a trivial solution, so let’s put this to one side.
Now you also quickly realise that if $f(x) > 0$ for some $x,\ $ then the following qualitative behaviour of the function $f(x)$ must be true: both $f(x)$ and $f’(x)$ must be strictly positive and increasing as $x {\to} \infty$. You realised this by drawing a small straight line of gradient 1 when $f(x) = 1$ and then $f’(x) = 1.1$ just to the right, when $f(x) = 1.1$ etc etc.
Are there any arguments- graphical, using calculus and continuity and any other tools you might have, to determine whether or not $f(c)=0 $ for some $ c \in \mathbb{R}$ as $x$ decreases? If such a $c$ does exist, then it follows that $f(x) = 0 \forall x<c$ also, since:
if $f(a) < 0$ for some $a<c$ then $\exists b $ with $a<b<c$ such that $f’(b) > 0$, so $f’(b) \neq f(b)$, a contradiction to the original condition. A similar argument can be made to show that $f(d) > 0$ for some $ d<c$ leads to a contradiction.
But it’s not clear to me how you can argue that the graph wouldn’t attain f(x) = 0 for some x. I don’t see why our graph can’t look like for example, the piecewise curve $y=0$ for $x < 0$ and $y= x^4$ for $x \geq 0$. Can an argument be made for why this isn’t possible this in a similar vein to one of my arguments/lines of reasoning above?
