# The function f'(x) = (x^3 -2)/x^4. If f(1) = 3, find f(x).

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### 2 Answers

We have f'(x) = (x^3 -2)/x^4.

To find f(x), we have to integrate f'(x) = (x^3 -2)/x^4

Int[ f'(x)] = Int[ (x^3 -2)/x^4]

=> Int[ 1/x] - Int[ 2/ x^4]

=> ln x - Int [ 2* x^-4]

=> ln x + (2/3)/x^3 + C

As f(1) = 3

=> ln 1 + (2/3)/1^3 + C = 3

=> 0 + 2/3 + C = 3

=> C = 7/3

**Therefore f(x) = ln x + (2/3)/x^3 + 7/3**

Given the derivative f'(x) = (x^3 -2) /x^4

We need to find f(x).

We know that the integral of f'(x) = f(x).

==> f(x) = intg (x^3-2)/x^4 dx

Let us simplify f'(x).

==> f(x) = intg (x^3/x^4) - 2/x^4 dx

= intg (1/x) - 2x^-4 dx

= intg (1/x)dx - intg (2x^-4) dx

= ln x - 2x^-3/-3 + C

= lnx + 2/3x^3 + C

But we know that f(1) = 3

==> f(1) = ln1 + 2/3 + C = 3

==> f(1) = 0 + 2/3 + C = 3

==> C = 3 - 2/3 = 7/3

**==> f(x) = lnx + 2/3x^3 + 7/3**