Use Newton's method to approximate the indicated root of the equation correct to six decimal places. The negative root of ex = 4 − x2

x = -1.964636

Step-by-step explanation:

Given equation;

eˣ = 4 - x²

This can be re-written as;

eˣ - 4 + x² = 0

Let

f(x) = eˣ - 4 + x²    -----------(i)

To use Newton's method, we need to get the first derivative of the above equation as follows;

f¹(x) = eˣ - 0 + 2x

f¹(x) = eˣ + 2x         -----------(ii)

The graph of f(x) has been attached to this response.

As shown in the graph, the curve intersects the x-axis twice - around x = -2 and x = 1. These are the approximate roots of the equation.

Since the question requires that we use the negative root, then we start using the Newton's law with a guess of x₀ = -2 at n=0

From Newton's method,

=> When n=0, the equation becomes;

Where f(-2) and f¹(-2) are found by plugging x = -2 into equations (i) and (ii) as follows;

f(-2) = e⁻² - 4 + (-2)²

f(-2) = e⁻² = 0.13533528323

And;

f¹(2) = e⁻² + 2(-2)

f¹(2) = e⁻² - 4 = -3.8646647167

Therefore

        [to 8 decimal places]

=> When n=1, the equation becomes;

Following the same procedure as above we have

=> When n=2, the equation becomes;

Following the same procedure as above we have

From the values of and , it can be seen that there is no change in the first 6 decimal places, therefore, it is safe to say that the value of the negative root of the equation is approximately  -1.964636 to 6 decimal places.

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step-by-step explanation:

ez do it yourself lazy