1. Help with antiderivatives?

If f'(x) = (-4x^2-2x)/(x^4) and f(1)=3, find f(x). what i did:
antiderivative[(-4 x x^-2)-(2 x x^-3)]dx
= (-4 x x^-1)-(2 x -1/2x^-2)]
= (4/x)+(1/x^2)
= (4/(1))+(1/(1)^2)
=5

i get 5 but shouldn't the answer be 3? what did i do wrong!

2. Originally Posted by yess
If f'(x) = (-4x^2-2x)/(x^4) and f(1)=3, find f(x). what i did:
antiderivative[(-4 x x^-2)-(2 x x^-3)]dx
= (-4 x x^-1)-(2 x -1/2x^-2)]
= (4/x)+(1/x^2)
= (4/(1))+(1/(1)^2)
=5

i get 5 but shouldn't the answer be 3? what did i do wrong!
$f'(x) = \frac{-4x^2-2x}{x^4}$

you forgot the constant of integration ...

$f(x) = \frac{4}{x} + \frac{1}{x^2} + C$

$3 = 4 + 1 + C$

$C= -2$

3. Originally Posted by skeeter
$f'(x) = \frac{-4x^2-2x}{x^4}$

you forgot the constant of integration ...

$f(x) = \frac{4}{x} + \frac{1}{x^2} + C$

$3 = 4 + 1 + C$

$C= -2$
ahh ok thank you!!

4. I get

$\displaystyle\int \frac{-4}{x^2}-\frac{2}{x^3}~dx = 4x^{-1}+2x^{-2}+C$

Then

$\displaystyle 3 = 4(1)^{-1}+2(1)^{-2}+C$

Can you finish it?

5. You should simplify, then use Partial Fractions...

$\displaystyle \frac{-4x^2-2x}{x^4} = \frac{-4x-2}{x^3}$.

You will need to use the decomposition of the form

$\displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}$.

6. Originally Posted by Prove It
You should simplify, then use Partial Fractions...

$\displaystyle \frac{-4x^2-2x}{x^4} = \frac{-4x-2}{x^3}$.

You will need to use the decomposition of the form

$\displaystyle \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x^3}$.
partial fractions?

$\displaystyle \frac{-4x^2-2x}{x^4} = -\frac{4x^2}{x^4} - \frac{2x}{x^4} = -\frac{4}{x^2} - \frac{2}{x^3}$.

7. Originally Posted by skeeter
partial fractions?

$\displaystyle \frac{-4x^2-2x}{x^4} = -\frac{4x^2}{x^4} - \frac{2x}{x^4} = -\frac{4}{x^2} - \frac{2}{x^3}$.
That way works too