Question

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

Answer 1

i cant find a way to isolate a or b to solve

Answer 2

have you found the derivative?

Answer 3

the derivative is 3ax^2 + 2bx but that doesnt help solve it

Answer 4

f(x) = ax^3 + bx^2

f(2) = 8a + 4b = -4

f(3) = 27a + 9b = 99

Answer 5

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.

Answer 6

Can you now solve for a, b ? @Lachlan1996

Answer 7

wait wait, hold on youve lost me here.

Answer 8

lol, take your time. Sorry :)

Answer 9

uhhh i got 27a + 6b =99 but that is the gradient as in the derivative

Answer 10

so i dont think you can simultaneously solve it

Answer 11

because you are simultaneously solving a derivative and a point

Answer 12

Yes! Now use the f(2)=-4 in the original equation to get your second equation with the two unknowns... a and b

Answer 13

Two equations with two unknowns should be solvable. Pick your favorite method for solving systems... @mathslover is using substitution.

Answer 14

but he is solving a f' (x) with a f (x)

Answer 15

a and b will be the same in the original and the derivative

Answer 16

ah yes i see

Answer 17

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.

Answer 18

in the above equation isnt it meant to be 6a not 9a though?

Answer 19

wait no i mean 6b not 9b

Answer 20

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

Answer 21

27a+6b=99
8a+ 4b=-4

Answer 22

uhh okie doke so a=0.5b-0.5

Answer 23

however substituting that into 27a + 6b = 99 i got 8.333333 as being b

Answer 24

I got a=7, b=-15

Answer 25

oh no made an error forgot to subtract 6b

Answer 26

OH MY FAULT , sorry :(

Answer 27

no, no it was a simple mistake, i just wasnt sure if you had made a mistake or were doing something else

Answer 28

this seems like alot of work just to start off the question, do you think this would be in an exam?

Answer 29

Depends but I do think, Yep! :)

Answer 30

For calculus, it doesn't seem out of bounds for a test.

Answer 31

ahh god, in a year 11 maths exam? i think im buggered then

Answer 32

Whether it will come or not in exam but I think you must be PREPARED

Answer 33

mmm yes my exam is in 5 weeks.

Answer 34

Creating a small system and solving it with substitution is fairly standard in calculus.

Answer 35

You have much time for it, go through these type of questions and ask your problems here...

Answer 36

okie doke, thanks very much for the help mate, its much appreciated. thankyou to you both

Answer 37

You're welcome.

Answer 38

Welcome! :)

Answer 39

uhhhh small problem

Answer 40

we got the answer wrong? its meant to be y=7x^3 -15x^2

Answer 41

Try to do it again, I think calculation mistake?

Answer 42

yes... a=7, b= -15

Answer 43

ah yes, last step of my working didnt put the -ve in the 15

Answer 44

hmn, Take care of these small mistakes .

Answer 45

whew! you scared me ;)

Answer 46

will do, doesnt help when you get frustrated but. maths is not easy for me

Answer 47

I think it is f'(3)=99 not f(3)=99