Well its asking for an x value that would satisfy that equation and as far as I could gather that's what I was trying to solve for in my calculations? I just need to figure it out before 1130 seeing as that's when my homework is due and this is the last problem I have
ok lol I am either very confused or there was some mis-communication. I'm using 1.5cos(x) as the function. so by that graph 1 would look like a logical choice therefor Newtons method reads x1-[f(x)/f'(x)] so I would get 1-[1.5cos(1)/(3sin(1)/2)] correct? If so that leaves me with a final value...
yes i understand how Newtons method works but I think the y=1.5cosx is confusing me.
I should be able to take any value for x and plug it in so let's say I take 3. Now that per Newtons method 3-[1.5cos(3)/(-3sin(3)/2)] this gives me -4.01525 and that is not correct which is why I'm stumped
ugghhh sorry I had a brain fart for a second it is positive infinity. thanks for all the help guys does anyone know how to complete the 2nd one I'm completely lost?? I relisted the problem with the correct info.
It would and your right it wouldn't even really need to apply hereI could just plug in zero. However if I do that I get a divided by 0 answer and I know the answer to the problem is a negative infinity so I'm not sure what's up.
The only thing is my problem is 3/(x^4) instead of 1/(x^4) I tried to apply hopitals rule to your example and I came out with just zero after taking the derivative of the result and then plugging in zero back into the derivative.
sorry it should read.
Use Newton's method to find the positive value of x which satisfies x=1.5cos(x) . Compute enough approximations so that your answer is within .05 of the exact answer. (You may use any starting point you deem appropriate.)
Hi all I have these problem on my test review and even after guessing one right I have no idea how one would arrive at that answer so your help is greatly appreciated.
Homework Statement
1.find the limit as x approaches 0 in the equation[(3/x^4)-(4/x^2)]
2.Use Newton's method to find the...
never mind lol I got it I used the quaratic on the top part of (4x^2+12x+2) and got the values and those ended up being my critical numbers. One thing though how come I didnt need to factor and solve the bottom? I thought that when a derivative was in the form of a fraction the top and bottom...
for the derivative per the quotient rule I got [(4x+6)(2x+1)-(4)(x^2+x+1)]/(x^2+x+1)^2 from there I multiplied out the top and got (4x^2+12x+2)/(x^2+x+1)^2 then I factored out a 2 from the top to give the equation i gave in my last post. I'm a little weary about factoring out that 2 though as in...
Thanks! I knew I had to plug it back in but for some reason I was getting the wrong answer when I plugged it in the equation the first time hahah. I managed to get all the other questions done except for this, I know I need to do the quotient rule but the equation I get for y' doesn't make any...
Homework Statement
Find the absolute maximum and absolute minimum values of the function
(x^3)+(12x^2)-27x+9
on each of the following interval
[-10,0]
The Attempt at a Solution
I got this and its saying its wrong.
y'= (3x^2)+(24x)-27
factor: 3(x^2+8x-9) so x=-9 and x=-1
y''=6x+24...