#### To determine

**To find**:

Root of the given function correct to 8 decimal places, using Newton’s Method.

#### Answer

1.0842246

#### Explanation

Given:fx=x2-x+1-4e-x2sinx.

Graph of f is shown below:

Since the graph of f cuts the x-axis, at exactly two points, therefore f has two zeros.

The Newton’s formula is:xn+1=xn-fxnf'xn+1.

f'x=2x-1-4(e-x2cosx-2x e-x2sinx). Let the initial approximation for first root be x0=0.2, then

x1=x0-fx0f'x0=0.2-f0.2f'0.2=0.21883273.

x2=x1-fx1f'x1=0.21883273-f0.21883273f'0.21883273=0.21916357.

x3=x2-fx2f'x2=0.21916357-f0.21916357f'0.21916357=0.21916368.

x4=x3-fx3f'x3=0.21916368-f0.21916368f'0.21916368=0.21916368.

Since x3=x4 upto 8 decimal places, first approximate root is 0.21916368.

As we have seen that there are exactly two zeros of f, so let’s consider approximation for second root, x0=1.1, then

x1=x0-fx0f'x0=1.1-f1.1f'1.1=1.0843283.

x2=x1-fx1f'x1=1.0843283-f1.0843283f'1.0843283=1.0842246.

x3=x2-fx2f'x2=1.0842246-f1.0842246f'1.0842246=1.0842246.

Since x2=x3 upto 8 decimal places, the approximate root is 1.0842246.

**Conclusion:**

We were able to find the root of the given function correct to 8 decimal places, using Newton’s Method.