**Intermediate Value Theorem**

Suppose that f(x) is continuous on [a, b] and let M be any number between f (a) and f (b). Then there exists a number C such that.

- a < c < b
- f (c) = M

All the intermediate value theorem is really saying is that a continuous functions will take on all values between f (a) and f (b). Below is a graph of a continuous function. That illustration the intermediate value theorem.

As we can see from this image if we pick any value, M, that is between the value of f (a) and the value of f (b) and draw a line straight out form this point the line will hit the graph in at least one point. In order words same where between a and b the function will take on the value of M. also, as the figure shows the function may take on the value at more than one place.

It’s also important to note that the intermediate value theorem only says that the function will take on the value of M. somewhere between a and b. it doesn’t say just what that value will be It only says that it exists.

**Example**: Show that f (x) = 2x³ – 5x² – 10 x + 5 has a root somewhere in the interval [-1, 2]

**Solution: **All we need to do is to show that the function is contains and that M = 0 is between f (-1) and f (2) i.e., f (-1) < 0 < f (2) Or f (2) < 0 < f (-1)

To do this all we need to is compute f (-1) = 8 and f (2) = -19.

So, we have

– 19 = f (2) < 0 < f (-1) = 8

∴ M = 0 is between f (-1) and f (2) and since f(x) is a polynomial it’s contains everywhere and so in particular it’s contains on the interval [-1, 2]. So, by the intermediate value theorem there must be a number -1 < c < 2 so that, f (c) = 0 therefore the polynomial does have a root between – 1 and 2.