How to use Newton's Method to solve an equation on an interval. Calculus 1 tutorial

Here we will use Newton's method to approximate the solution to the quartic equation 3x^4-8x^3+2=0 on the interval [2, 3]. This is one of the most common ways to approximate the roots of an equation and the idea is based on the equation of tangent lines.

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[maximeclere]

4:58 XD the answer has 2.7 and 3.14 in it

[WerewolfLord]

4:55 The answer is approximately music plays*]
*it doesn't but it should

[tejarex]

Newton's method works great when it works, but it fails when the initial guess is not good enough, which is pretty easy in general. Given a continuous function f and 2 points where f has opposite signs, finding an intermediate root by bisection, always keeping two points with opposite signs, is guaranteed to work. For this example, assuming that the 2.63 root is unique, one would check 2.5, 2.75, 2.625, 2.6875, 2.65625, etc.

[bradley]

So the answer was closer to 3 anyway? That's interesting. Good video ty

[robertveith]

@ blackpenredpen -- I don't see your rationale for picking x_1 to be 3. I would have averaged the endpoint values and used 2.5 instead.

[kepler]

now do it with euler's method now

[kiit]

Where is the full lecture for newton method?? Help plz

[Tiwey]

Isnt this the differentiating quotient of the normal?

[gatocomcirrose]

i mean, if you have to do this by hand it's just better to use wolfram alpha

[TristanANain]

Thank you sir for the lesson but the sad face tho😂

[robertveith]

3x^3: x = 8/3 - 2/(3x^3). If you let x_1 = 3, then x_2 ~ 2.6419753, x_3 ~ 2.6305154, ..., x_6 ~
2.6300203