Find the image and preimage of functions Given the function $f(x)=e^{(x-1)^2}, f:\mathbb{R}  → \mathbb{R}$ ,  find $f[[0, +∞)] $.
Given the function $f(x)=2x^2 + 5x - 5, f:\mathbb{R}→\mathbb{R}$ ,  find $f^{-1} [[0, 3]] $.
I have no idea how to solve the first example, so I would be very grateful for explanation. 
In the following example I got $f^{-1} [[0, 3]] = [-28,\frac{-5 - \sqrt{65}}{4}]\cup [\frac{-5 + \sqrt{65}}{4}, 28] $. I am not sure if I'm right. Could you tell me if my result is correct?
 A: For the first function observe that it is a composition of $(x-1)^2$ and $e^x$ and since both are continuous and moreover, they are differentiable on $\Bbb R$, the composition i.e. $e^{(x-1)^2}$ is differentiable over $\Bbb R$. 
So differentiating, $f^/(x) = 2(x-1)e^{(x-1)^2}$ and $$ f^{(2)}(x) = {4(x-1)^2}e^{(x-1)^2} + 2e^{(x-1)^2}$$.
Now equating, $f^/(x) = 0 $ we obtain that x = 1.
Putting x = 1 at  $ f^{(2)}(x) = {4(x-1)^2}e^{(x-1)^2} + 2e^{(x-1)^2}$ we deduce that $f^{(2)}(x) > 0$, so f has a global minima at x = 1 . At that point , f(1) = 1.
So at all other points $f(x)>1$ and since $f$ is continous on a connected interval, it satisfies the IVP and thus answer to your first question becomes $[1,+\infty)$ .
[Actually, you can consider the graph of $e^{(x-1)^2}$ to be that of $e^{x^2}$ with the origin right-shifted at x = 1.]
For the second question observe that $2x^2 + 5x - 5$ is a polynomial and so it is also continuous and also since the set you're considering the pre-image of is a connected set you just need to solve the quadratic for the boundary points of the inetrval i.e. you just need to solve 
$$  2x^2 + 5x - 5 = 0$$ and $$2x^2 + 5x - 5 = 3$$ 
The roots you'll obtain probably, $$x = \frac{-5 \pm \sqrt{65}}{4}$$ for the first one and $$x = \frac{-5 \pm \sqrt{89}}{4}$$ for the second one. Now take suitable union and you'll obtain answer . 
[$\frac{-5 + \sqrt{89}}{4}$ will replace 28 in your answer and $\frac{-5 - \sqrt{89}}{4}$ will replace -28]
A: For the first example, calculate $f(0)$, see if the function is increasing. If it is, then the image is $[f(0), \infty]$, the function is continuous and explodes to infinit.
For the second one, solve the inequation $0 \leq 2x^{2}+5x−5 \leq 3$. Wolfram says that you are wrong. Calculate $f(28)$ you are going to see that you are wrong.
