If `f(x)=x(1-x)^(3)` then which of following is true ?

Options

A. f(x) has local maxima at x=1
B. f(x) has local minima at x=1
C. f(x) has local minima at ` x=(1)/(4)`
D. f(x) has local maxima at ` x=(1)/(4)`

Answer

To solve the problem, we need to analyze the function \( f(x) = x(1-x)^3 \) and determine the nature of its critical points, specifically whether they correspond to local maxima, local minima, or points of inflection.

### Step 1: Find the first derivative \( f'(x) \)

Using the product rule for differentiation, we have:

\[ f'(x) = \frac{d}{dx}[x] \cdot (1-x)^3 + x \cdot \frac{d}{dx}[(1-x)^3] \]

Calculating the derivatives:

1. The derivative of \( x \) is \( 1 \).
2. The derivative of \( (1-x)^3 \) using the chain rule is \( -3(1-x)^2 \).

Thus, we can write:

\[ f'(x) = (1)(1-x)^3 + x(-3(1-x)^2) \]

Simplifying this expression:

\[ f'(x) = (1-x)^3 - 3x(1-x)^2 \]

Factoring out \( (1-x)^2 \):

\[ f'(x) = (1-x)^2 \left( (1-x) - 3x \right) \]

This simplifies to:

\[ f'(x) = (1-x)^2 (1 - 4x) \]

### Step 2: Set the first derivative to zero

To find the critical points, we set \( f'(x) = 0 \):

\[ (1-x)^2 (1 - 4x) = 0 \]

This gives us two cases:

1. \( (1-x)^2 = 0 \) implies \( x = 1 \)
2. \( 1 - 4x = 0 \) implies \( x = \frac{1}{4} \)

Thus, the critical points are \( x = 1 \) and \( x = \frac{1}{4} \).

### Step 3: Find the second derivative \( f''(x) \)

Now, we need to find the second derivative to determine the nature of the critical points. We differentiate \( f'(x) \):

Using the product rule again on \( f'(x) = (1-x)^2 (1-4x) \):

\[ f''(x) = \frac{d}{dx}[(1-x)^2] \cdot (1-4x) + (1-x)^2 \cdot \frac{d}{dx}[(1-4x)] \]

Calculating the derivatives:

1. The derivative of \( (1-x)^2 \) is \( -2(1-x) \).
2. The derivative of \( (1-4x) \) is \( -4 \).

Thus, we have:

\[ f''(x) = -2(1-x)(1-4x) + (1-x)^2(-4) \]

Simplifying this expression:

\[ f''(x) = -2(1-x)(1-4x) - 4(1-x)^2 \]

Factoring out \( (1-x) \):

\[ f''(x) = (1-x) \left[-2(1-4x) - 4(1-x)\right] \]

Expanding the terms inside the brackets:

\[ = (1-x) \left[-2 + 8x - 4 + 4x\right] \]
\[ = (1-x) \left[12x - 6\right] \]

### Step 4: Evaluate the second derivative at the critical points

1. **At \( x = 1 \)**:

\[ f''(1) = (1-1)(12 \cdot 1 - 6) = 0 \]

This indicates that \( x = 1 \) is a point of inflection.

2. **At \( x = \frac{1}{4} \)**:

\[ f''\left(\frac{1}{4}\right) = \left(1 - \frac{1}{4}\right)\left[12 \cdot \frac{1}{4} - 6\right] \]
\[ = \left(\frac{3}{4}\right)\left[3 - 6\right] = \left(\frac{3}{4}\right)(-3) = -\frac{9}{4} \]

Since \( f''\left(\frac{1}{4}\right) < 0 \), this indicates that \( x = \frac{1}{4} \) is a local maximum.

### Conclusion

The correct statement is:

- \( f(x) \) has a local maximum at \( x = \frac{1}{4} \).