Question

Find the positive root of x3-6x+4=0 by newton raphson method.

Answer 1

Answer:

The positive root is 0.73.

Step-by-step explanation:

Let, the function f(x) = $x^{3} - 6x + 4$.

We need to find the value of xn for which f(xn) = 0, by newton raphson method.

Lets take x=0.

$\frac{d f(x)}{dx} = 3x^{2} - 6$.

f(0) = 4 and at x = 0, $\frac{d f(x)}{dx} = -6$.

Hence, next value of x will be $0 - \frac{4}{-6} = \frac{2}{3}$.

Now, f($\frac{2}{3}$) = 0.2962 (approx 0).

So, lets take a further value of x that is x2.

f($\frac{2}{3}$) = 0.2962 and $\frac{d f(\frac{2}{3} )}{dx} = 3(\frac{2}{3} )^2 - 6 = -\frac{14}{3}$.

Thus, x2 = $\frac{2}{3} - \frac{0.2962}{-\frac{14}{3} } = 0.73$.

f(0.73) = 0.008409 ≅ 0.

Answer 2

Step-by-step explanation:

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