# Intermediate Value Theorem

The intermediate value theorem states that if 'f' is continuous on a closed interval [a,b], f(a) $\neq$ f(b) and 'k' is any number between f(a) and f(b) , then there is at least one number 'c' in [a,b] such thatf(c) = k.

This theorem is used to

1) prove that the roots of the given polynomial exists.

2) There are two points diametrically opposite each other on the surface of the Earth where the temperature are exactly same.

3) Climatologist use intermediate value theorem to predict the level of carbon di-oxide which affect the Earth.

## Examples on Intermediate Value Theorem

1) Use the intermediate-value theorem to show that the polynomial f(x)= $x^{3}$ + 2x -1 has a zero in the interval of [0,1].**Solution :**As the given function is continuous in the closed interval [0,1]

f(0) = $0^{3}$ + 2(0) -1 = -1 and

f(1) = $1^{3}$ + 2(1) -1 = 2

From the above we can see that

f(0) < 0 and f(1) > 0 which means there exist 'c' in [0,1] such that

f(c) = 0 From the above graph we can see that f(x) is continuous in the closed interval [0,1].

2) Apply intermediate-value theorem and find the value of 'c' in the interval of [0,5] for the function f(x) = $x^{2}$ + x -1 and f(c) = 11.

**Solution :**As we know that 'f' is continuous in the given interval [0,5] so there exists some 'k' between f(0) and f(5) such that

f(c) = k

As f(c) 11 so, k = 11, a = 0 and b = 5

Now we will find the value of the function at these points.

f(0) = $0^{2}$ + 0 -1 = -1

f(5) = $5^{2}$ + 5 -1 = 29

So by intermediate-value theorem

f(a) < k < f(b)

-1 < k < 29

Given that f(c) = 11

$c^{2}$ + c -1 = 11

$c^{2}$ + c -12 = 0

(c + 4)(c - 3) = 0

So, c = -4 or c = 3

As the given interval is [0,5] So the value of c = 3.

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