#### To determine

**To find**:

Root of the given function correct to 6 decimal places, using Newton’s Method.

#### Answer

The approximate root is 1.050064.

#### Explanation

**Given**:

ln(4-x2)-x.

Graph of f is shown below:

Since the graph of f cuts the x-axis, at exactly two points, -1.9, 1.0 therefore f has two zeros.

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

f'x=(-2x)4-x2-1. . Let the initial approximation for first root be x0=-1.99, then

x1=x0-fx0f'x0=-1.9975.

x2=x1-fx1f'x1=-1.9674.

x3=x2-fx2f'x2=-1.9647528.

x4=x3-fx3f'x3=-1.9646358.

x5=x4-fx4f'x4=-1.9646355.

Since x5=x4 upto 6 decimal places, first approximate root is -1.964635.

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=1.0586.

x2=x1-fx1f'x1=1.058007.

x3=x2-fx2f'x2=1.058006401.

x4=x3-fx3f'x3=1.058006401.

Since x4=x3 upto 6 decimal places, the approximate root is 1.050064.

**Conclusion:**

The approximate root is 1.050064.