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anonymous
 4 years ago
given that f(x) = ax^3 + bx^2, f(2)=4 and f'(3)=99 find f(x), f(3) and f'(2)
anonymous
 4 years ago
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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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i cant find a way to isolate a or b to solve

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0have you found the derivative?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the derivative is 3ax^2 + 2bx but that doesnt help solve it

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2f(x) = ax^3 + bx^2 f(2) = 8a + 4b = 4 f(3) = 27a + 9b = 99

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.28a + 4b = 4 27a+9b=99 from equation 1 , we have : a = \(\LARGE\frac{44b}{8}=\frac{1b}{2}\) put this value in the second equation.

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2Can you now solve for a, b ? @Lachlan1996

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait wait, hold on youve lost me here.

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2lol, take your time. Sorry :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0uhhh i got 27a + 6b =99 but that is the gradient as in the derivative

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so i dont think you can simultaneously solve it

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0because you are simultaneously solving a derivative and a point

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes! Now use the f(2)=4 in the original equation to get your second equation with the two unknowns... a and b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Two equations with two unknowns should be solvable. Pick your favorite method for solving systems... @mathslover is using substitution.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but he is solving a f' (x) with a f (x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0a and b will be the same in the original and the derivative

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2f(2) = 8a + 4b = 4 f(3) = 27a + 9b = 99 a = (1b)/2 put this in f(3) equation. 27(1b)/2 + 9b = 99 (2727b+18b)=198 27 9b = 198 9b = 198+27 = 225 b = 225/9 b = 25 You can solve for a now.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0in the above equation isnt it meant to be 6a not 9a though?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait no i mean 6b not 9b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Its 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
 4 years ago
Best ResponseYou've already chosen the best response.0uhh okie doke so a=0.5b0.5

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0however substituting that into 27a + 6b = 99 i got 8.333333 as being b

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh no made an error forgot to subtract 6b

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2OH MY FAULT , sorry :(

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0no, no it was a simple mistake, i just wasnt sure if you had made a mistake or were doing something else

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this seems like alot of work just to start off the question, do you think this would be in an exam?

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2Depends but I do think, Yep! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For calculus, it doesn't seem out of bounds for a test.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ahh god, in a year 11 maths exam? i think im buggered then

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2Whether it will come or not in exam but I think you must be PREPARED

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mmm yes my exam is in 5 weeks.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Creating a small system and solving it with substitution is fairly standard in calculus.

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2You have much time for it, go through these type of questions and ask your problems here...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okie doke, thanks very much for the help mate, its much appreciated. thankyou to you both

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we got the answer wrong? its meant to be y=7x^3 15x^2

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2Try to do it again, I think calculation mistake?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah yes, last step of my working didnt put the ve in the 15

mathslover
 4 years ago
Best ResponseYou've already chosen the best response.2hmn, Take care of these small mistakes .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0whew! you scared me ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0will do, doesnt help when you get frustrated but. maths is not easy for me

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think it is f'(3)=99 not f(3)=99
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