**Intermediate value theorem.**

Let be continuous. Suppose for , . If c is a real number between and , then there is an between and such that .

**Proof.**

Define by . Then, g is a continuous function such that and are of opposite signs. The assertion of the theorem amounts to saying that there is a point between and such that . Without loss of generality, we may take and (otherwise replace g by -g). If , we write and ; otherwise, write and so that we have , . Now if , then .

If , write and , otherwise write and , so that we have and . We could continue this process and find sequences ,

with and and

,

.

Since is a monotonically non-decreasing sequence bounded above, it must converge. Suppose it converges to . Similarly, is monotonically non-increasing, bounded below and therefore converges to, say, . We further note that as implying . Let us call this . By the continuity of g, we have , and since for all n, we must have and at the same time since for all n, we must also have . This implies . QED.

**Corollary. **

If f is a continuous function in an interval I and for some , then there is a point c between a and b for which . (**Exercise**).

*The above result is often used to locate the roots of equations of the form *.

For example, consider the equation: .

Note that whereas . This shows that the above equation has a root between 0 and 1. Now try with 0.5. . So there must be a root of the equation between 1 and 0.5. Try . , which means that the root is between 0.5 and 0.75. So, we may try . . So the root is between 0.75 and 0.625. Now, if we take the approximate root to be 0.6875, then we are away from the exact root at most a distance of 0.0625. If we continue this process further, we shall get better and better approximations to the root of the equation.

**Exercise. **

Find the cube root of 10 using the above method correct to 4 places of decimal.

More later,

Nalin Pithwa