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?
Best Answer
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.