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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?

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Perhaps more to the point, do you understand what is being asked for in this question, or is that where you’re stuck? –  Brian M. Scott Nov 18 '12 at 19:25
    
I know it is continuous throughout that interval. But, mathematically I do not know how to verify it. –  AFerrara Nov 18 '12 at 19:26
2  
You verify it as follows: First, note that $f(8) = 3$ and $f(35) = 6$. Then let $y$ be any real number in $[3,6]$, and show that there exists $x \in [8,35]$ such that $f(x)=y$. In shis case, $x$ can be expressed as a simple function of $y$. –  TonyK Nov 18 '12 at 19:33

3 Answers 3

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.

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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.

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$$(1)\;\;\;\;f(8)=\sqrt 9=3\;\;,\;\;f(35)=\sqrt{36}=6$$

$$(2)\;\;\;\;\forall\,c\in (3,6)\,\,\exists\,x_c\in (8,35)\,\,s.t.\,\,f(x_c)=c\Longleftrightarrow \sqrt {x_c+1}=c\Longleftrightarrow x_c+1=c^2$$

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.

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