I need to prove with the Intermediate Value Theorem that $\sqrt s$ exists, where $s > 0$.
My textbook states this definition of the Intermediate Value Theorem: Suppose that $f$ is continuous on the closed interval $[a,b]$ and let $N$ be any number between $f(a)$ and $f(b)$, where $f(a)$ is not equal to $f(b)$. Then there exists a number $c$ in $(a,b)$ such that $f(c) = N$.
Some values were already given to me:
- $f(x) = x^2$
- $a = 0$
- $b = s + (1/2)$
- Using the above values, I get $f(a) = 0$ and $f(b) = (s+(1/2))^2$
So I'm assuming that I have to choose a value for c that fits the criteria? So if I choose $c = s$, that would be in $(a,b)$ and $f(c) = c^2 = s^2$ would be between $f(a)$ and $f(b)$. But does that really prove anything? I guess I just don't understand how the Intermediate Value Thoerem proves the existence of a square root.
Best Answer
You should start with the function $f(x)=x^2-s$. Then $f(0)=-s <0$ and $f\left(s+\frac{1}{2}\right)=\left(s+\frac{1}{2}\right)^2-s=s^2+\frac{1}{4} >0$. Now use intermediate value theorem to claim that $f$ can be $0$.