bad initial approximation for Newton's Method

When we use Newton's Method to approximate the roots of a function (or the zeros of an equation), we must be careful that our initial guess does not have its derivative being zero.

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#justcalculus

[volodymyrgandzhuk]

If f(x)=x³-3x+6, then, as it has been said in the video, f'(x)=3x²-3. And if we set 3x²-3=0, we will find two values for x. One of them is 1, this has also been said in the video; the other one is -1. So, f(x) has two stationary points: -1 and +1.
f(-1)=-1+3+6=8>0
f(1)=1-3+6=4>0
This means that our equation only has one real solution, and this solution is less than -1.
I will start by plugging in -2 and then only writing out the solutions at each step:
x_0=-2
x_1≈-2.4444
x_2≈-2.3592
x_3≈-2.3553
x_4≈--2.3553
So, the solution of our equation has an approximate value of -2.3553.

[hypergraphic]

Thanks for putting out all these videos. Really appreciate it.

[shubhadeepshome]

x = 1 turned out to be one of the roots of f'(x)! As the slope at the extrema is zero, we can't apply Newton's Method because we'd be dividing by 0.

[bprpcalculusbasics]

To see a worked-out example of Newton's method,

[saharhaimyaccov]

Can u explain how newton found this formula?

[SJrad]

X=1 is a local minimum of that function

[darkwing]

Is that the only case that could be a bad initial approximation (values that make the derivative 0)?

[kiit]

Why do u hold that pokeball😁, , , which pokemon is inside... Char