anonymous
  • anonymous
given that f(x) = ax^3 + bx^2, f(2)=-4 and f'(3)=99 find f(x), f(3) and f'(2)
Mathematics
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
i cant find a way to isolate a or b to solve
anonymous
  • anonymous
have you found the derivative?
anonymous
  • anonymous
the derivative is 3ax^2 + 2bx but that doesnt help solve it

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More answers

mathslover
  • mathslover
f(x) = ax^3 + bx^2 f(2) = 8a + 4b = -4 f(3) = 27a + 9b = 99
mathslover
  • mathslover
8a + 4b = -4 27a+9b=99 from equation 1 , we have : a = \(\LARGE\frac{-4-4b}{8}=\frac{-1-b}{2}\) put this value in the second equation.
mathslover
  • mathslover
Can you now solve for a, b ? @Lachlan1996
anonymous
  • anonymous
wait wait, hold on youve lost me here.
mathslover
  • mathslover
lol, take your time. Sorry :)
anonymous
  • anonymous
uhhh i got 27a + 6b =99 but that is the gradient as in the derivative
anonymous
  • anonymous
so i dont think you can simultaneously solve it
anonymous
  • anonymous
because you are simultaneously solving a derivative and a point
anonymous
  • anonymous
Yes! Now use the f(2)=-4 in the original equation to get your second equation with the two unknowns... a and b
anonymous
  • anonymous
Two equations with two unknowns should be solvable. Pick your favorite method for solving systems... @mathslover is using substitution.
anonymous
  • anonymous
but he is solving a f' (x) with a f (x)
anonymous
  • anonymous
a and b will be the same in the original and the derivative
anonymous
  • anonymous
ah yes i see
mathslover
  • mathslover
f(2) = 8a + 4b = -4 f(3) = 27a + 9b = 99 a = (-1-b)/2 put this in f(3) equation. 27(-1-b)/2 + 9b = 99 (-27-27b+18b)=198 -27 -9b = 198 -9b = 198+27 = 225 b = -225/9 b = -25 You can solve for a now.
anonymous
  • anonymous
in the above equation isnt it meant to be 6a not 9a though?
anonymous
  • anonymous
wait no i mean 6b not 9b
anonymous
  • anonymous
Its actually a really cool way to come up with a system... one eq from the original and one from the derivative. Slick, huh?! correct... I got 6b rather than 9b as well
anonymous
  • anonymous
27a+6b=99 8a+ 4b=-4
anonymous
  • anonymous
uhh okie doke so a=0.5b-0.5
anonymous
  • anonymous
however substituting that into 27a + 6b = 99 i got 8.333333 as being b
anonymous
  • anonymous
I got a=7, b=-15
anonymous
  • anonymous
oh no made an error forgot to subtract 6b
mathslover
  • mathslover
OH MY FAULT , sorry :(
anonymous
  • anonymous
no, no it was a simple mistake, i just wasnt sure if you had made a mistake or were doing something else
anonymous
  • anonymous
this seems like alot of work just to start off the question, do you think this would be in an exam?
mathslover
  • mathslover
Depends but I do think, Yep! :)
anonymous
  • anonymous
For calculus, it doesn't seem out of bounds for a test.
anonymous
  • anonymous
ahh god, in a year 11 maths exam? i think im buggered then
mathslover
  • mathslover
Whether it will come or not in exam but I think you must be PREPARED
anonymous
  • anonymous
mmm yes my exam is in 5 weeks.
anonymous
  • anonymous
Creating a small system and solving it with substitution is fairly standard in calculus.
mathslover
  • mathslover
You have much time for it, go through these type of questions and ask your problems here...
anonymous
  • anonymous
okie doke, thanks very much for the help mate, its much appreciated. thankyou to you both
anonymous
  • anonymous
You're welcome.
mathslover
  • mathslover
Welcome! :)
anonymous
  • anonymous
uhhhh small problem
anonymous
  • anonymous
we got the answer wrong? its meant to be y=7x^3 -15x^2
mathslover
  • mathslover
Try to do it again, I think calculation mistake?
anonymous
  • anonymous
yes... a=7, b= -15
anonymous
  • anonymous
ah yes, last step of my working didnt put the -ve in the 15
mathslover
  • mathslover
hmn, Take care of these small mistakes .
anonymous
  • anonymous
whew! you scared me ;)
anonymous
  • anonymous
will do, doesnt help when you get frustrated but. maths is not easy for me
anonymous
  • anonymous
I think it is f'(3)=99 not f(3)=99

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