#### Intermediate Value Theorem :

- If a function f is continuous on closed interval [a, b] and f (a). f (b) < 0, i.e. ƒ (a) and ƒ (b) are opposite signs, then there exists at least one c ∈ (a, b) such that ƒ(c) = 0.
- This theorem is to be used without proof. We know that a polynomial function is continuous and so above theorem can be applied to a polynomial function. Hence there exists at least one root of polynomial equation ƒ(x) = 0, in the open interval (a, b).
- We will be studying three methods of approximation of roots of equation ƒ (x) = 0.

### Method – I : Bisection Method

**Example – 1:**

- Using the bisection method solve the equation x
^{2}+ 2x – 8 = 0 in the interval [1, 4]. Use two iterations.

**Solution:**

Let ƒ(x) = x^{2} + 2x – 8

∴ ƒ(1) = (1)^{2} + 2(1) – 8 = 1 + 2 – 8 = -5 < 0 (negative)

∴ ƒ(4) = (4)^{2} + 2(4) – 8 = 16 + 8 – 8 = 16 > 0 (positive)

Thus we see that ƒ(1) < 0 and ƒ(4) > 0

∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 4) = (a, b)

**Initial approximation : (First approximation)**

Let x_{o} be the initial (first) approximation to the root.

By the bisection formula

x_{o} = (a + b)/2 = (1 + 4)/2 = 5/2 = 2.5

**First iteration (Second approximation):**

Now, ƒ(2.5) = (2.5)^{2} + 2(2.5) – 8

ƒ(2.5) = 6.25 + 5 – 8

ƒ(2.5) = 3.25 > 0 (positive)

Now, ƒ(1) < 0 and ƒ(2.5) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 2.5) = (a, b).

Let x_{1} be the second approximation to the root.

By the bisection formula

x_{1} = (a + b)/2 = (1 + 2.5)/2 = 3.5/2 = 1.75

**Second iteration (Third approximation):**

Now, ƒ(1.75) = (1.75)^{2} + 2(1.75) – 8

= 3.0635 + 3.5 – 8

= -1.4375 < 0 (negative)

Now, ƒ(1.75) < 0 and ƒ(2.5) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.75, 2.5) = (a, b)

Let x_{2} be the third approximation to the root.

By the bisection formula

x_{1} = (a + b)/2 = (1.75 + 2.5)/2 = 4.25/2 = 2.125

After two iteration by bisection method solution of equation x^{2} + 2x – 8 = 0 is x = 2.125. (approx.)

#### Example – 02:

- Show that the root of the equation x2 + 3x – 5 = 0 lies in (1,2). Find the first three approximations to the roots of this equation using the bisection method.

**Solution :**

Let ƒ(x) = x^{2} + 3x – 5

∴ ƒ(1) = (1)^{2} + 3(1) – 5 = 1 + 3 – 5 = -1 < 0 (negative)

∴ ƒ(2) = (2)^{2} + 3(2) – 5 = 4 + 6 – 5 = 5 > 0 (positive)

Thus we see that ƒ(1) < 0 and ƒ(2) > 0

∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 2) = (a, b)

**Initial approximation : (First approximation)**

Let x_{0} be the initial approximation (first approximation) to the root.

By the bisection formula

x_{0} = (a + b)/2 = (1 + 2)/2 = 3/2 = 1.5

**First iteration (Second approximation):**

Now, ƒ(1.5) = (1.5)^{2} + 3(1.5) – 5

= 2.25 + 4.5 – 5

= 1.75 > 0 (positive)

Now, ƒ(1) < 0 and ƒ1.5) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 1.5) = (a, b)

Let x_{1} be the second approximation to the root.

By the bisection formula

x_{1} = (a + b)/2 = (1 + 1.5)/2 = 2.5/2 = 1.25

**Second iteration (Third approximation):**

Now, ƒ(1.25) = (1.25)^{2} + 3(1.25) – 5

= 1.5625 + 3.75 – 5

= 0.3125 > 0 (positive)

Now, ƒ(1) < 0 and ƒ(1.25) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 1.25) = (a, b)

Let x_{2} be the third approximation to the root.

By the bisection formula

x2 = (a + b)/2 = (1 + 1.25)/2 = 2.25/2 = 1.125

Thus the first three approximations to the root of equation x^{2} + 3x – 5 = 0 by bisection method are 1.5, 1.25 and 1.125

#### Example – 3:

- Show that the root of the equation x
^{3}– x – 1 = 0 lies in (1,2). Find the first three approximations to the roots of this equation using the bisection method.

**Solution:**

Let f (x) = x^{3} – x – 1

∴ ƒ(1) = (1)^{3} – (1) – 1 = 1 –1 -1 = -1 < 0 (negative)

∴ ƒ(2) = (2)^{3} – (2) – 1 = 8 – 2 – 1 = 5 > 0 (positive)

Thus, we see that ƒ(1) < 0 and ƒ(2) > 0

∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 2) = (a, b)

**Initial approximation : (First approximation)**

Let x_{o} be the initial approximation (first approximation) to the root.

By the bisection formula

x_{o} = (a + b)/2 = (1 + 2)/2 = 3/2 = 1.5

**Second approximation :(First iteration)**

Now, ƒ(1.5) = (1.5)^{3} – 1.5 – 1

= 3.375 – 1.5 –1

= 0.875 > 0 (positive)

Now, ƒ(1) < 0 and ƒ(1.5) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1, 1.5) = (a, b)

Let x_{1} be the second approximation to the root.

By the bisection formula

x1 = (a + b)/2 = (1 + 1.5)/2 = 2.5/2 = 1.25

**Third approximation :(Second iteration)**

Now, ƒ(1.25) = (1.25)^{3} – 1.25 – 1

= 1.953125 – 1.25 – 1

= – 0.296875 < 0 (negative)

Now, ƒ(1.25) < 0 and ƒ(1.5) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.25, 1.5) = (a, b).

Let x_{2} be the third approximation to the root.

By bisection formula,

x_{2} = (a + b)/2 = (1.25 + 1.5)/2 = 2.75/2 = 1.375

Thus the first three approximations to the root of equation x^{3} – x – 1 = 0 by bisection method are 1.5, 1.25 and 1.375.

#### Example – 4:

- Using the bisection method find the approximate value of square root of 3 in the interval (1, 2) by performing two iterations.

**Solution:**

Let x = √3,

Squaring both the sides we have,

x^{2} = 3

∴ x^{2} – 3 = 0

The positive root of this equation is √3.

Let f(x) = X2 – 3

∴ ƒ(1) = (1)2 – 3 = 1 – 3 = -2 < 0 (negative)

∴ ƒ(2) = (2)2 – 3 = 4 – 3 = 1 > 0 (positive)

∴ By intermediate value theorem, the root of equation ƒ(x) = 0 lies in interval (1, 2) = (a, b)

**Initial approximation : (First approximation)**

Let x_{0} be the initial approximation (first approximation) to the root.

By the bisection formula

x_{0} = (a + b)/2 = (1 + 2)/2 = 3/2 = 1.5

**First Iteration (Second approximation) :**

Now, ƒ(1.5) = (1.5)^{2} – 3

= 2.25 – 3

= – 0.75 < 0 (negative)

Now, ƒ(1.5) < 0 and ƒ(2) > 0, hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.5, 2) = (a, b)

Let x_{1} be the second approximation to the root.

By bisection formula

x_{1} = (a + b)/2 = (1.5 + 2)/2 = 3.5/2 = 1.75

**Second Iteration (Third approximation) :**

Now, ƒ(1.75) = (1.75)^{2} – 3

= 3.0625 – 3

= 0.0625 > 0 (positive)

Now, ƒ(1.5) < 0 and ƒ(1.75) > 0 and , hence by intermediate value theorem the root of equation ƒ(x) = 0 lies in interval (1.5, 1.75) = (a, b)

Let, x_{2} be the third approximation to the root.

By bisection formula,

x_{2} = (a + b)/2 = (1.5 + 1.75)/2 = 3.25/2 = 1.625

After two iterations by the Bisection method √3 = 1.625

Example – 05:

- Find the square root of 6 using the bisection method and two iterations, (Ans. 2.375)

Example – 06:

- Find the fourth root of 27 using the bisection method and two iterations, (Ans. 2.375)

Example =- 07:

- Show that the root of equation x3 – 2×2 + 2 = 0 in the interval (-1, 0) by using bisection method three times (Ans. – 0.875)