Antiderivative?

Question

f'(x)= (8/x^2) - (4/x^6)

fins f(x) when f(1)=0

Answer 1

Since you have f' = a/x^n you can use the reverse of the rule for differentiation

d(1/x^n)/dx = -n/x(n+1)

So consider f' = 8/x^2 - 4/x^6

For teh first term n+1 = 2 which means that n = 1. We should have a factor of (-1) in the numerator but since we don't, there has to be a factor of (-1/1) multiplying the term. So the anti-derivative of the first term is

-8/x

YOu can go through the same thought process for the second term and find

4/5*1/x^5 so that

f(x) = -8/x +4/5*1/x^5+ c where c is an arbitraty constant

Now f(1) = -8+4/5+c = 0 ----> c = 8 -4/5 =36/5

so f(x) = -8/x +4/5 1/x^5 + 36/5

Answer 2

f'(x)=(8/x^2)-(4/x^6)

f(x)=(8x^-2)-(4x^-6)

=(8*(x^-2+1)/-2+1) - (4*(x^-6+1)/-6+1)

=(8 *x^-3/-3)-(4*x^-7/-7)

f(x)=(8/-3x^3)-(4/-7x^7)+c


if f(1)=0 then:

f(1)=(8/-3(1^3))-(4/-7(1^7))+c=0

=(8/-3)-(4/-7)+c=0

= -40/21+c=0

=c=40/21

therefore:f(x)=(8/-3x^3)-(4/-7x^7)+40/21

Answer 3

(-8/(x^3)) + (4/(5x^5)) + C

-8 + 4 + C = 0
C = 4

(-8/(x^3)) + (4/(5x^5)) + 4