f''(x) = 6x + sinx , Find f

Question

anonymous · Answer

i dont know what to do with the 6x

zzr0ck3r · Answer

what is the derivative of 3x^2?

anonymous · Answer

do we not work with antiderivatives?

zzr0ck3r · Answer

you do, the derivative os 3x^2 = 6x so the anti derivative of 6x = 3x^2

anonymous · Answer

3x^2 is an antiderivative of 6x

zzr0ck3r · Answer

the anti derivative of a*x^n = a/(n+1) * x^(n+1)

zzr0ck3r · Answer

so do the opposite, increment the exponent by 1, then devide the coefficent by that number

anonymous · Answer

so the answer is x3-sinx+c ?

anonymous · Answer

x^3 - sinx +c

anonymous · Answer

Is the problem given the first or second derivative?  I can't see....

anonymous · Answer

it's the second

anonymous · Answer

Oh.. ok.. then i shouldn't have erased... :(

anonymous · Answer

The first time you took the anti derivative you'll get the first derivative f'.  So you should have gotten f'(x) = 3x^2 - cosx + C1.

To get f, you'll have to take the anti derivative again...
So f(x) = x^3/6 - sinx + C1x + C2

Try taking the derivative twice to see that you have the correct second derivativ....
$\large f(x)=\frac{x^3}{6}-sinx+C_1x+C_2 $    where C1 and C2 are constants.

anonymous · Answer

isn't the antiderivative of 3x^2 = 3x^3/3 = x^3?

anonymous · Answer

Yes it is...

anonymous · Answer

so where did the x^3/6 derive from?

anonymous · Answer

Oh... lemme check...

anonymous · Answer

$\large f(x)=x^3-sinx+C_1x+C_2 $

you're right... :)
I think I'm too stuffed with turkey....

anonymous · Answer

haha Happy Thanksgiving!

anonymous · Answer

Same to u and everyone else!