`(d)/(dx) (AC)= f(x) (MC-AC)`, then f(x)= (i) `(1)/(x)` (ii) x (iii) `x^(-2)` (iv) `-x^(-1)`

Options

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

Answer

To solve the equation \(\frac{d}{dx} (AC) = f(x) (MC - AC)\), we need to find the function \(f(x)\).

### Step-by-Step Solution:

1. **Understanding Average Cost (AC)**:
The average cost \(AC\) is defined as the total cost \(C\) divided by the quantity \(x\):
\[ AC = \frac{C}{x} \]

2. **Differentiating Average Cost**:
We need to differentiate \(AC\) with respect to \(x\):
\[ \frac{d}{dx}(AC) = \frac{d}{dx}\left(\frac{C}{x}\right) \]

3. **Applying the Quotient Rule**:
To differentiate \(\frac{C}{x}\), we use the quotient rule, which states that if \(u = C\) and \(v = x\), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, \(u = C\) and \(v = x\), so:
\[ \frac{d}{dx}\left(\frac{C}{x}\right) = \frac{x \frac{dC}{dx} - C \cdot 1}{x^2} \]
This simplifies to:
\[ \frac{d}{dx}(AC) = \frac{x \frac{dC}{dx} - C}{x^2} \]

4. **Identifying Marginal Cost (MC)**:
The marginal cost \(MC\) is defined as:
\[ MC = \frac{dC}{dx} \]
Substituting this into our equation gives:
\[ \frac{d}{dx}(AC) = \frac{x \cdot MC - C}{x^2} \]

5. **Rearranging the Expression**:
We can factor out \(\frac{1}{x}\):
\[ \frac{d}{dx}(AC) = \frac{1}{x} \left(MC - \frac{C}{x}\right) \]
Recognizing that \(\frac{C}{x} = AC\), we rewrite it as:
\[ \frac{d}{dx}(AC) = \frac{1}{x} (MC - AC) \]

6. **Comparing with Given Equation**:
We have:
\[ \frac{d}{dx}(AC) = f(x)(MC - AC) \]
Setting this equal to our derived expression:
\[ f(x)(MC - AC) = \frac{1}{x}(MC - AC) \]

7. **Solving for \(f(x)\)**:
To find \(f(x)\), we can equate the coefficients:
\[ f(x) = \frac{1}{x} \]

### Conclusion:
Thus, the function \(f(x)\) is:
\[ f(x) = \frac{1}{x} \]