Question

If \(\dfrac{d}{dx}f(x) = 3x^2 - \dfrac{3}{x^4}\), and \(f(1) = 0\), then \(f(x)\) is:

• A. \(6x + \dfrac{12}{x^5}\)
• B. \(x^4 - \dfrac{1}{x^3} + 2\)
• C. \(x^3 + \dfrac{1}{x^3} - 2\)
• D. \(x^3 + \dfrac{1}{x^3} + 2\)

Hint:

Always integrate term by term. Use known power rules: \[ \int x^n dx = \frac{x^{n+1}}{n+1} + C \quad (n \neq -1) \] Then apply the given initial condition to find the constant of integration.

Answer 1

The Correct Option is C

We are given: \[ \frac{d}{dx}f(x) = 3x^2 - \frac{3}{x^4} \] To find \(f(x)\), we integrate both terms with respect to \(x\): \[ f(x) = \int \left(3x^2 - \frac{3}{x^4} \right) dx = \int 3x^2 \, dx - \int \frac{3}{x^4} \, dx \] \[ = 3 \cdot \frac{x^3}{3} - 3 \cdot \frac{x^{-3}}{-3} + C = x^3 + \frac{1}{x^3} + C \] Now, use the initial condition \(f(1) = 0\) to find the constant \(C\): \[ f(1) = 1^3 + \frac{1}{1^3} + C = 1 + 1 + C = 2 + C = 0 \Rightarrow C = -2 \] Therefore, \[ f(x) = x^3 + \frac{1}{x^3} - 2 \] But this corresponds to Option (C) — yet you've marked the correct answer as (D). Let us double-check: \[ f(x) = \int (3x^2 - \frac{3}{x^4}) dx = x^3 + \frac{1}{x^3} + C \] With \(f(1) = 0\), we find: \[ 0 = 1 + 1 + C \Rightarrow C = -2 \] \[ \boxed{f(x) = x^3 + \frac{1}{x^3} - 2} \]