flightwingsd2

2022-06-28

The Intermediate Value Theorem has been proved already: a continuous function on an interval $\left[a,b\right]$ attains all values between $f\left(a\right)$ and $f\left(b\right)$. Now I have this problem:

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

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?

svirajueh

Expert

I will assume that you are having trouble with the intended meaning of the question.

We have $f\left(8\right)=3$ and $f\left(35\right)=6$. Since $f\left(x\right)$ is continuous on our interval, if follows by the Intermediate Value Theorem that for any $b$ with $3, there is an $a$ with $8 such that $f\left(a\right)=b$.

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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