How does one verify the Intermediate Value Theorem? The Intermediate Value Theorem has been proved already: a continuous function on an interval $[a,b]$ attains all values between $f(a)$ and $f(b)$. Now I have this problem:

Verify the Intermediate Value Theorem if $f(x) = \sqrt{x+1}$ in the interval is $[8,35]$.

I know that the given function is continuous throughout that interval. But, mathematically, I do not know how to verify the theorem. What should be done here? 
 A: I will assume that you are having trouble with the intended meaning of the question.  
We have $f(8)=3$ and $f(35)=6$. Since $f(x)$ is continuous on our interval, if follows by the Intermediate Value Theorem that for any $b$ with $3\lt b\lt 6$, there is an $a$ with $8\lt a \lt 35$ such that $f(a)=b$.
You are being asked to show that the Intermediate Value Theorem holds in this specific situation without using the IVT. Effectively, you are being asked to express the required $a$ in terms of $b$, and to verify that it is indeed between $8$ and $35$.
So we want $\sqrt{a+1}=b$. Now you can take over. 
A: The function $f(x)$ is strictly increasing on $[8,35]$, $f(8)=3$, and $f(35)=6$, so what you’re being asked to show is that for every $y\in[3,6]$ there is an $x\in[8,35]$ such that $f(x)=y$. Suppose that someone handed you a number $y$ between $3$ and $6$; could you tell him how to find an $x$ such that $\sqrt{x+1}=y$? (If that’s not enough, mouse-over the spoiler protected bit below for a further hint.)

 HINT: Inverse function.

A: $$(1)\;\;\;\;f(8)=\sqrt 9=3\;\;,\;\;f(35)=\sqrt{36}=6$$
$$(2)\;\;\;\;\forall\,c\in (3,6)\,\,\exists\,x_c\in (8,35)\,\,s.t.\,\,f(x_c)=c\Longleftrightarrow \sqrt {x_c+1}=c\Longleftrightarrow x_c+1=c^2$$
and since $\,c\in (3,6)\Longleftrightarrow x_c+1=c^2\in (9,36)\,\Longrightarrow x_c\in (8,35)$ , and we're done.
