#1

Hey the question is out of a trial hsc paper:

The graph of y=f(x) passes through the point (2.12) and f'(x)=9x^2 + 4. Find f(x).

I think you need to use a formual with k in it but i forget what it is and i forgot my textbook at skool lol. A cookie for whoever can help Thanks.

The graph of y=f(x) passes through the point (2.12) and f'(x)=9x^2 + 4. Find f(x).

I think you need to use a formual with k in it but i forget what it is and i forgot my textbook at skool lol. A cookie for whoever can help Thanks.

#2

You would have to integrate the f'(x), to get f(x), I think.

#3

so f '(x) passes through (2,12) as well? find f(x)?

integrate f '(x)

f(x) = 3x^3 +4x +c

then plug in the x value (2) and the y value (12)

12 = 3(2)^3 + 4(2) + c

c =-20? i did this in my head sorry

now pug -20 back in for the integral constant and you have your f(x).

(goes to sleep)

integrate f '(x)

f(x) = 3x^3 +4x +c

then plug in the x value (2) and the y value (12)

12 = 3(2)^3 + 4(2) + c

c =-20? i did this in my head sorry

now pug -20 back in for the integral constant and you have your f(x).

(goes to sleep)

*Last edited by LawnDwarf at Jul 24, 2008,*

#4

The graph of y=f(x) passes through the point (2.12) andf'(x)=9x^2 + 4. Find f(x).

isn't that f(x)?

i dont do advanced maths yet though, ive only just left school, so i could be wrong...

#5

Integrate f'(x), then use the given coordinate to work out the value of c (the constant term)

EDIT: Beaten to it

EDIT: Beaten to it

*Last edited by National_Anthem at Jul 24, 2008,*

#6

ok, integrating f'(x) gives you:

y = 3x^3 + 4x + c.

Now, you have the point (2,12), so sub to find c.

12 = 3*8 + 4*2 + c

So

12 = 32 + c

c = -20

Hence, f(x) = 3x^3 + 4x - 20.

I think thats it, anyhoo.

y = 3x^3 + 4x + c.

Now, you have the point (2,12), so sub to find c.

12 = 3*8 + 4*2 + c

So

12 = 32 + c

c = -20

Hence, f(x) = 3x^3 + 4x - 20.

I think thats it, anyhoo.

#7

the bold there is actualy F prime of x

#8

Hey the question is out of a trial hsc paper:

The graph of y=f(x)passes through the point (2.12) andf'(x)=9x^2 + 4.Find f(x).

I think you need to use a formual with k in it but i forget what it is and i forgot my textbook at skool lol. A cookie for whoever can help Thanks.

FOUND IT!!! not only once but twice!!

#9

isn't that f(x)?

i dont do advanced maths yet though, ive only just left school, so i could be wrong...

It could be f'(x) or f(x), but they have stated that it is f'(x) in this case.

so f '(x) passes through (2,12) as well? find f(x)?

integrate f '(x)

f(x) = 3x^3 +4x +c

then plug in the x value (2) and the y value (12)

12 = 3(2)^3 + 4(2) + c

c =20? i did this in my head sorry

now pug 20 back in for the integral constant and you have your f(x).

(goes to sleep)

...and this is right.

#10

ohh ok i got confused by the question lol thanks to all who helped, i can be at ease ... until i find another question i cant do ...

#11

FOUND IT!!! not only once but twice!!

Fail, f'(x) is not f(x).