Introduction

Welcome back, you made it to the last article of Unit 1! We will now talk about another notable theorem, the Intermediate Value Theorem (IVT). This theorem shows up often on the test, so make sure you know what it is and how to apply it.

IVT states: If a function $f(x)$ is continuous on an interval $[a,b]$ and a value $L$ exists between $f(a)$ and $f(b)$ then there exists some $c$ in the interval $[a,b]$ such that $f(c)=L$. 

This means that if there is a continuous function on an interval, that function’s value on that interval will contain all the values between the function’s values at the endpoints of that interval.

Now that you know what IVT is, let’s jump into some example problems in case you are still confused

Examples

Example 1: Show that if $f(3)=-2.8$ and $f(5)=3.2$, the function $f(x)$ has at least one zero given that $f(x)$ is a continuous function on all real numbers.

$f(x)$ being continuous on all real numbers means it will be continuous on the interval $[3,5]$. Next, $0$ is between $f(3)$ and $f(5)$, as in $f(3)<0<f(5)$. Therefore, there exists some value $c$ in the interval $[3,5]$ such that $f(c)=0$. 

Not only were we able to deduce that the function had a zero, we were also able to figure out that it lies somewhere inside the interval.

One thing to note about the IVT is that it is an existence theorem, which means that other things can still happen. The function could also have zeroes outside the interval $[3,5]$, or there could be a value of $f(x)$ greater than $5$ or less than $-3$. However, we only know for certain that there is at least one zero on the interval $[3,5]$. 

Example 2: Given that $g(x)$ is continuous and $g(-2)=g(8)=10$, is there guaranteed to be a value of c such that $g(c)=6$ by IVT?

No there isn’t. The number $6$ is not between $g(-2)$ and $g(8)$ as they both equal $10$. IVT cannot be applied here. There could be a value of $c$ such that $g(c)=6$, but the IVT does not guarantee that.

Example 3: Given that $h(2)=27$ and $h(4)=36$ is there a value of $c$ in the interval $[2,4]$ such that $h(c)=30$ due to the IVT?

No, it is unknown whether $h(x)$ is continuous, so the IVT cannot be applied here.

Now that we’ve covered a few different types of problems, we can jump into the practice section.

 Practice