0
$\begingroup$

Prove that $\exists$ a unique differentiable function $f: R \to R$ s.t. $f'(x) = e^{x^2} \forall x \in R$ and $f(0) = 0$

I'm having trouble showing the uniqueness part. This is what I have so far. Write $F(x) = \int_{a}^{x} e^{t^2} dt$

Choose $a = 0$

$F(x) = \int_{0}^{x} e^{t^2} dt$

We are told $F(0) = 0$, so

$F(0) = 0 = \int_{0}^{0} e^{t^2} dt = 0$

So the integral exists. Now we must show it is unique. I am really not sure where to go from here. I think the $e^{x^2}$ is throwing me off because there is no anti-derivative. So I'm not sure how I would show it is unique here. Any help is appreciated!

$\endgroup$

1 Answer 1

2
$\begingroup$

Assume $G$ is another such function. Then $H:=G-F$ is differentiable with $H'\equiv 0$ and $H(0)=0$. Then $H(x)=0$ for all $x$. Indeed, by the Mean Value Theorem, there exists $\xi $ between $0$ and $x$ such that $\frac{H(x)-H(0}{x-0}=H'(\xi)=0$.

$\endgroup$
2
  • $\begingroup$ So because the new function (H) still needed the same conditions, it means it is the same function? Because it kind of looks like you just found another function that works. $\endgroup$
    – Nolan P
    Dec 10, 2020 at 13:07
  • $\begingroup$ I see that you edited it. However, I'm still not sure I understand. It seems like you found a function H that satisfies the conditions. So it seems to me this shows it is not unique. Can you try explaining more? I am still confused. $\endgroup$
    – Nolan P
    Dec 13, 2020 at 14:54

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .