given that f(x) = ax^3 + bx^2, f(2)=-4 and f'(3)=99 find f(x), f(3) and f'(2)

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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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i cant find a way to isolate a or b to solve
have you found the derivative?
the derivative is 3ax^2 + 2bx but that doesnt help solve it

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

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