How To Prove This Is Continuous Or Discontinuous How do I prove this is continuous/discontinuous?
A function $f(x)$ (can be any function, in general) is defined on the interval $[3,7]$, where $f(3)=2$ and $f(7)=4$. The range of $f(x)$ on this interval is $$2\leq f(x)\leq 4$$ for all $x \in [3,7]$.
$f$ has unique values on $[2,4]$ (so if $b$ is in $[2,4]$ there is only one $c$ in $[3,7]$ such that $f(c)=b$)
How should I prove that this is continuous/not continuous? I think using the intermediate value theory, we can assume that f(x) is continuous first and then use say that there is a point $c$ and  $$f(c) = \alpha$$ and $f(a) &lt; \alpha &lt; f(b)$. What do I do with the point next? The function $f(x)$ can be anything so I cannot use a single value for it. Please can someone help.
If you can prove that for every alpha between f(a) and f(b) the function takes on a value, then you have proved the intermediate value theorem. In this cases, what I look for is a counter example. What you want is to look for all the f(c). That would be tedious. Just look for a counter example, such that f(d) = $\beta$ and $\beta$ does not lie in the interval between f(a) and f(b). If you cannot find such d, then from the intermediate value theorem, it is continuous.
If you can prove that for every alpha between f(a) and f(b) the function takes on a value, then you have proved the intermediate value theorem. In this cases, what I look for is a counter example. What you want is to look for all the f(c). That would be tedious. Just look for a counter example, such that f(d) = $\beta$ and $\beta$ does not lie in the interval between f(a) and f(b). If you cannot find such d, then from the intermediate value theorem, it is continuous.