Section 2.3 Intermediate Values
Recall the graph of this function from the first section of this chapter. 
Notice that it has the value \(0\) at \(0\) (i.e. \(f(0)=0\)) and it has the value \(1.5\) at \(1\) (i.e. \(f(1)=1.5\)). But, does it ever have the value \(1.3\) (i.e. is there a domain value \(x\) for which \(f(x)=1.3\))? Imagine a horizontal line on the graph at a height of \(1.3\) and you’ll see that the answer is "no". So, why not? What goes wrong that this function has the value \(0\) and \(1.5\) but never not has the intermediate value \(1.3\text{?}\) Hopefully the answer is apparent. Namely, the function has a discontinuity so there is a "break" / "jump" over some of the intermediate values like \(1.3\text{.}\) One of the nice properties of functions that are continuous on an interval of numbers (i.e. continous at every number in an interval) is that it obtains all of its intermediate values. Specifically, we have the following theorem known by mathematicians as the "Intermediate Value Theorem (IVT)".
Consider for example the function above but this time defined with \(f(1)=1\text{.}\) Here, let’s plot it. 
Now, if you look at any two values the function obtains, you’ll see that every value in between them is also obtained by the function. In other words, point one finger at a number on the vertical axis that the function obtains and also point a finger at another number on the vertical axis that the function obtains and then consider any number in between. Look over at the graph and you’ll see that the function also obtains that value. Why? Well, we removed the disontinuity so now this function is continuous on the interval \(I=[0,2]\) so we have the IVT.
