use newton's method to estimate the requested solution of the equation. start with given value of Xo and then give X2 as the estimated solution. 3x^2+2x-1=0; Xo = 1; Find the right-hand solution.

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use newton's method to estimate the requested solution of the equation. start with given value of Xo and then give X2 as the estimated solution. 3x^2+2x-1=0; Xo = 1; Find the right-hand solution.

Mathematics
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get the derivative, so you know the slope of the line at x0 = 1
find the value of f(1) so you have a set of coordinates to use for your "equation of the slope" line to find the new "x1" with.
f(1) = 4; (1,4) we will use this to calibrate our line equation with; now we find the slope at f(1)

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i see
f'(x) = 6x +2 f'(1) = 8 ; our slope is 8 y = mx+b 4 = 8(1) + b 4 = 8 + (-4) y = 8x -4 find the root :) the x int and that your "new" X value
with y = to 1 right
when y = 0 we have our x intercept, (x,0) is the x int..
0 = 8x -4 x = 4/8 = 1/2
right. man your awesome
to keep punching along, we use the NEW x value in our original equation and keep getting closer and closer to our "true" root ;) or farther away depending on the nature of the curve
i see. hey is there a way to find the critical points of a function on a calculator instead of using calculus
calculus is the "method" of the math used to find this stuff. calculus is applied to the situation when nothing else can get us there :) So I would say that it depends on the calculator and the functions that it has. But in the end, they rely upon the principles of calculus to get the answer
true. well maybe since ur smart you can help with a calc problem i have.
the other maths are really applied calculus :) tell me, is the earth round or flat?
i might be able to help....my smarts come and go lol
calculus, in essense, says that in order to figure out a problem you ahve to get close enough to it to "flatten" things out to where we can use our innate understanding of the "flat" world around us :)
Interesting we'll hv to chat more abt that... i'm trying to Find the function with the given derivative whose graph passes through the point P. r'(t) = sec^2t - 4, P(0,0)
that would be an integral. or antiderivative.. same thing different jargon. [S] sec^2(t) - 4 dt or \[\int\limits_{}\sec^{2} - 4 dt\]
just like dressing down into the derivative, we just suit up into the integral... they are opposite movements
[S] sec^2(t) dt -4 [S] dt does this make sense?
lol not really. the answer i came up with is sec t - 4t - 4
what function do you know that dress down to a derivative of sec^2? what function do you know that dress down to a derivative of 1 ?
x^2
no
x
lets check that:....or it looks like you already did
x dresses down to 1 d(1x) = 1 that is correct. So what do we dress "1" to be? x right? we just back out of the derivative into the original function right?
-4 [S] dt -> -4t now for the left part :) sec^2 came from what? do you recall your trig dervivatives?
I think: d(tan(x)) = sec^2 did I recall correctly?
i think ur right
so would the answer be tan t - 4t
I think im right to, but Ive been known to be mistaken ;) so: [S] sec^2(t) dt suits back up into tan(t) our original equation is of the form: f(x) = tan(t) -4t but that aint it exactly...that just gives us a "family" of curves to choose from...we need to pin this down with a constant
tell me, can 2 different equations have the exact same derivative? y = 2x^2 + 6 y = 2x^2 -8
yes
right?
that right: so what we have right now is an equation to a curve that is actually floating up and downthe y axis and we need to pin it down with something....... we do that with a constant (C). like this: y = tan(t) -4t +C
does that make sense?
so thats the answer
that is getting to our answer, remember they gave us an initial condition that the curve passes thru the point (0,0) right? so lets use this information to find the actaul value for "C".
ok
P(0,0) 0 = tan(0) -4(0) + C solve for C :)
C would b zero
thats right, but now we "know" for sure :) so the equation is: r = tan(t) -4t and were done :)
i like how you make sure i know where the answer is coming from. you would make a great instructor
thanx; I figure that if I show you whats going on under the problem that you might just be able to use it in another place ;)
i'm working on this word problem. "From a thin piece of cardboard 40in. by 40in., square corners are cut oout so that the sides can be folded up to make a box. What dimensions will yield a box of maximum volume? what is the maximum volume rounded to nearest tenth if necessary
like this right?
1 Attachment
Volume is to maximized so we need the formula for the volume of a box: Do you know what that would be?
length x width x height, right?
thats correct; when we cut out these corners and fold them up, that is our "height" right? lets call that x..ok?
length and width are equal measurements given the problem, 40 and 40 right? when we remove "x" from on end and "x" from the other; that makes our sides shorter by x+x. length then becomes: 40- 2x and width becomes: 40-2x lets use this in our formula for volume: V = (40-2x)(40-2x)(x) we agree?
ok
we can factor out x right
factor out? now, we need to multiply all these together to find the equation for derive :)
so far our box is like this:
1 Attachment
ok
V = (40-2x)(40-2x)(x) V = 1600x -160x^2 +4x^3 you wanna derive this for me?
sure i got 1600 - 320x + 12x^2
thats right... now we need to equal that to 0 to find our max and min.... for x :) 4(3x^2 -80x +400) = 0 3x^2 -80x +400 = 0
this gives me 2 numbers: x = 40/3 and x = 20/3
both of these numbers should give us an equal volume of box, but different shapes.
unless I am mistaken, and I think I am with that last one...
plug both of these into our forula stuff and see which is bigger :)
which formula
V = (40-2(40/3)) (40-2(40/3)) (40/2) V = (40-2(20/3)) (40-2(20/3)) (20/3) we only have one formula for volume that needs a value for x ;)
20/3 yields a bigger answer
for 40/3 we get 2014.81 .....then 20/3 it is :)
yep, for 20/3 we get 7407.41
i got 4740 for 20/3
yeah, I noticed that I punched in the wrong thing in the calculator :)
so the bigger # is the volume?
yep....when we cut out a corner of "20/3" from each side, we have gotten the maximum volume we can get out of it
cool

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