If $\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$ such that $f(2) = 0$. Then $f(x)$ is (A) $x^4 + \frac{1}{x^3} + \frac{129}{8}$ (B) $x^3 + \frac{1}{x^4} + \frac{129}{8}$ (C) $x^4 + \frac{1}{x^3} - \frac{129}{8}$ (D) $x^3 + \frac{1}{x^4} - \frac{129}{8}$

Question

Accepted Answer

> [!TLDR]
> We are given the derivative $f'(x) = 4x^3 - \frac{3}{x^4}$ and the initial condition $f(2)=0$. Integrating term by term gives $f(x) = x^4 + \frac{1}{x^3} + C$, and using $f(2)=0$ yields $C = -\frac{129}{8}$. So $f(x) = x^4 + \frac{1}{x^3} - \frac{129}{8}$, which matches option **(C)**.

This is a classic **Initial Value Problem** (IVP). You are given the rate of change of a function (its derivative) and one specific value of the function itself. The idea is simple: if you know how fast something is changing and you know where it started, you can reconstruct the whole story. Here, the derivative tells us the slope of $f$ at every $x$, and the point $f(2)=0$ anchors the curve at exactly one spot. Integrating the derivative recovers $f$ up to an unknown constant; the initial condition pins that constant down.

Let’s walk through it.

1. **Integrate the derivative.**  
   We have $f'(x) = 4x^3 - \frac{3}{x^4}$.  
   Rewrite $\frac{3}{x^4}$ as $3x^{-4}$ to make the power rule clear.  
   Then
   $$f(x) = \int \left(4x^3 - 3x^{-4}\right) dx.$$
   Integrate term by term:
   - $\int 4x^3 \, dx = 4 \cdot \frac{x^{4}}{4} = x^4$,
   - $\int -3x^{-4} \, dx = -3 \cdot \frac{x^{-3}}{-3} = x^{-3} = \frac{1}{x^3}$.
   
   So
   $$f(x) = x^4 + \frac{1}{x^3} + C,$$
   where $C$ is the constant of integration.

> [!TIP]
   > A common slip is forgetting the sign when integrating $x^{-4}$. The rule $\int x^n dx = \frac{x^{n+1}}{n+1}$ works for $n=-4$: $\frac{x^{-3}}{-3}$, and then multiplying by $-3$ gives $+x^{-3}$. Always check by differentiating: the derivative of $\frac{1}{x^3}$ is $-\frac{3}{x^4}$, which matches the given term.

2. **Use the initial condition to find $C$.**  
   We know $f(2) = 0$. Substitute $x=2$ into the expression: …

Q.If $\frac{d}{dx} f(x) = 4x^3 - \frac{3}{x^4}$ such that $f(2) = 0$ . Then $f(x)$ is (A) $x^4 + \frac{1}{x^3} + \frac{129}{8}$ (B) $x^3 + \frac{1}{x^4} + \frac{129}{8}$ (C) $x^4 + \frac{1}{x^3} - \frac{129}{8}$ (D) $x^3 + \frac{1}{x^4} - \frac{129}{8}$

Concept understanding — Initial Value Problem

Initial Value Problem

The intuition: a rule of change plus a starting point

The precise statement

A worked example

Method: Solve an initial-value problem (integrate, then fix the constant)

Steps

Common Mistakes

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