Range of a function from real numbers to real numbers I am finding the range of the following function
$$f: R \rightarrow R \\ f(x)=(x+1)^2+1$$
From the definition, I got that $$\forall_{x \in R}   f(x) \ge 1 $$
It is obvious to notice that since $x \in R$, x takes every possible real number. From the definition we can say that $\forall_{x \in R} f(x)  \ge 1 \implies$ range of $f \subset [1,\infty ) $.
But how can I prove the fact that range of $f = [1,\infty )  $ i.e., $(x+1)^2+1$ produces every possible real number greater than or equal to 1.
 A: Suppose $y\in [1, \infty)$; you need to show that there is some $x\in\mathbb{R}$ with $f(x)=y$.
To do this, you'll have to use preestablished facts about the functions involved (or prove them now - but I'm assuming you've already proved them, or seen them proved). Specifically, you argue roughly as follows:


*

*Since $y\in[1, \infty)$, $y\ge 1$; so $y-1\ge 0$.

*Any number $\ge 0$ has a square root (this is a fact you should already know), so there is some $z\in\mathbb{R}$ such that $z^2=y-1$.

*We can subtract $1$ from any real number - so, subtracting $1$ from $z$, we get that there is some $a\in\mathbb{R}$ such that $(a+1)^2=y-1$.
Do you see how to go from here? (HINT: what can you say about $f(a)$?)
A: Let $y$ be in the range of the function. Then
$$(x+1)^2+1=y$$
$$(x+1)^2=y-1$$
$$x=-1\pm\sqrt{y-1}.$$
Then for any value of $y\ge 1$, you get a corresponding value of $x$.
Now suppose that $y<1$. Then $y-1$ is negative, so you cannot have $(x+1)^2=y-1$, since the left side is a nonnegative number.
A: You can do it by producing an inverse: Given any $y \in [1,\infty)$, pick $x = \sqrt{y-1}-1$; verify that $x$ is well-defined and that $f(x) = y$.
A: Try providing an explicit expression for $x$ that will give you any desired $y$.  
For example, if you claim that $f(x) = 15$ for some $x$, then why don't you show me the value of $x$? Now do it for the general case with your constraints.  
