Strictly monotone increasing Let $f$ be a real-valued function defined on an interval $[a, \infty)$. One says that $f(x)$
tends to infinity with $x$ if for every $B \gt 0$ there is an $A \gt 0$ such that for every
$x \gt A$ we have $f(x) \gt B$.
Let $f$ be a strictly monotone increasing, continuous function on an interval $[a, \infty)$.
Set $c = f(a)$, and assume that $f(x)$ tends to infinity with $x$. Using the intermediate value theorem, prove that for every $y$ in $[c, \infty)$ there exists one and only one
$x$ in $[a, b)$ such that $f(x) = y$.
Can some help me with this, explain for the capital $A$ and $B$ Thanks
 A: Intuitively, "$f(x)$ tends to infinity with $x$"  means that as $x$ gets large, $f(x)$ gets larger than any number you pick.  Because $f(x)$ is monotonic, once it gets larger than something, it stays larger than that something.  
The definition is how to prove it.  Say you claim that $f(x)$ goes to infinity.  I am allowed to challenge you with some number $B$.  To meet the challenge you have to respond with some number $A$ that guarantees if $x \gt A,$ then $f(x) \gt B$.  This is a translation of the $\epsilon - \delta$ proof of a limit to "numbers near $\infty$.
For two examples, let $f(x)=\frac x{1000}$  If I challenge you with $B=1,000,000$, you can answer $A=1,000,000,000$ and demonstrate that if $x \gt 10^9, f(x) \gt 10^6$.  You can even show that whatever $B$ I name, you win if $A=1000B$.  Let $g(x)=10^9-\frac 1x$.  If I challenge you with $B=10^6$, you can win with any $A \gt 0$.  But if I am clever enough to challenge you with $B=10^{12}$ you lose.  So $g(x)$ doesn't go to $\infty$ with $x$.
You might look at Jared's answer to the problem you were given.
Added:  Your questioner has told you that $f(x)$ goes to infinity with $x$, given you a $y \gt f(a)$ and asked you to find an $x$ such that $f(x)=y$ and prove it unique.  To prove it exists, challenge him with $B=2y$ and he will give you an $A$ such that if $z \gt A, f(z) \gt 2y$.  Now the intermediate value theorem shows there is such an $x$ in $[a,A]$ and monotonicity shows it is unique.
A: Edit: it's only after writing all this that I realized Jared had done pretty much the same thing...sorry.
Let $y\geq f(a)$. Since $\lim_{+\infty} f=+\infty$, there exists $A>0$ such that $f(x)>y$ for all $x>A$. Note that this forces $A> a$. Now set $b=A+1$. You get $f(b)>y\geq f(a)$. By the intermediate value theorem, there exists $x\in[a,b]$ such that $f(x)=y$.
It remains to prove uniqueness of such an $x$ above. Take $y\geq f(a)$. Assume this is attained for two distinct values. That is $a\leq x_1<x_2$ are such that $y=f(x_1)=f(x_2)$. Since $f$ is increasing, we have $f(x_1)<f(x_2)$. Contradiction.
So for every $y\in [f(a),+\infty)$, there exists a unique $x\in[a,+\infty)$ such that $f(x)=y$.
Note: the result we've just proved says exactly that $f:[a,+\infty)\longrightarrow[f(a),+\infty)$ is bijective.
A: Choose some $y\in[c,\infty)$.  Since $f(x)$ tends to infinity as $x$ does, this means there is some $b$ with $f(b)> y$.  Thus we have a continuous function $f(x)$, a number $y$ in the closed interval $[a,b]$, and $f(a)\le y< f(b).$  The intermediate value theorem guarantees the existence of some $x\in[a,b)$ with $f(x)=y$.
To show uniqueness, suppose there were another $x_0$ satisfying $f(x_0)=y$.  Since either $x<x_0$ or $x_0< x$, this would violate the condition that $f(x)$ is strictly monotone.  Hence, $x$ is unique.
