Please solve the following problem by using newton raphson method?

Question

Where do the curves of $y = cos x$ $y = cosx$ and $y = x 3 - 1$ $y = x3 −1$ intersect?

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Answer 1 · 2018-04-27T10:59:47

$x \approx 1.12656$ $x approx 1.12656$

Consider the graphs of $y = cos x and y = x^{3} - 1$ $y=cosx and y=x^3-1$ below:
graph{(y-cosx)(y-x^3+1)=0 [-10, 10, -5, 5]}

We can see that the graphs intersect at some point greater than $x = 1$ $x=1$

The intersection point occurs where: $cos x = x^{3} - 1$ $cosx = x^3-1$

That is where: $cos x - x^{3} + 1 = 0$ $cosx-x^3+1=0$

The Newton/Raphson iteration method says that for $f (x) = 0$ $f(x)=0$

$x_(i+1) = x_i - (f(x_i))/(f'(x_i))$

Where, $x_(i+1)$ is a better estimate of $x$ than $x_i$

In this case: $f(x) = cosx-x^3+1$
and $f'(x)= -sinx-3x^2$

We will take $x_0 =1$ (from the graphs above).

We can then iterate as follows:

enter image source here
Hence, $x=approx 1.12656$