The function `f(x)=x^(5)-5x^(4)+5x^(3)` has

Options

A. One minima and two maxima
B. Two minima and one maxima
C. Two minima and two maxima
D. One minima and one maxima

Explanation

To analyze the function \( f(x) = x^5 - 5x^4 + 5x^3 \) for its maxima and minima, we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
To find the critical points where the function may have maxima or minima, we first need to compute the first derivative of the function.

\[ f'(x) = \frac{d}{dx}(x^5 - 5x^4 + 5x^3) \]
Calculating the derivative term by term:
\[ f'(x) = 5x^4 - 20x^3 + 15x^2 \]

### Step 2: Set the first derivative to zero
Next, we set the first derivative equal to zero to find the critical points.

\[ 5x^4 - 20x^3 + 15x^2 = 0 \]
Factoring out the common term \( 5x^2 \):
\[ 5x^2(x^2 - 4x + 3) = 0 \]
This gives us:
\[ x^2 = 0 \quad \text{or} \quad x^2 - 4x + 3 = 0 \]

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation \( x^2 - 4x + 3 = 0 \) using factoring:
\[ (x - 1)(x - 3) = 0 \]
Thus, the roots are:
\[ x = 1 \quad \text{and} \quad x = 3 \]
Including the root from \( 5x^2 = 0 \), we have the critical points:
\[ x = 0, \quad x = 1, \quad x = 3 \]

### Step 4: Find the second derivative \( f''(x) \)
To determine whether these critical points are maxima or minima, we need to compute the second derivative.

\[ f''(x) = \frac{d}{dx}(5x^4 - 20x^3 + 15x^2) \]
Calculating the second derivative:
\[ f''(x) = 20x^3 - 60x^2 + 30 \]

### Step 5: Evaluate the second derivative at the critical points
Now we will evaluate \( f''(x) \) at each critical point.

1. **At \( x = 0 \)**:
\[ f''(0) = 20(0)^3 - 60(0)^2 + 30 = 30 \quad (\text{positive, hence minimum}) \]

2. **At \( x = 1 \)**:
\[ f''(1) = 20(1)^3 - 60(1)^2 + 30 = 20 - 60 + 30 = -10 \quad (\text{negative, hence maximum}) \]

3. **At \( x = 3 \)**:
\[ f''(3) = 20(3)^3 - 60(3)^2 + 30 = 540 - 540 + 30 = 30 \quad (\text{positive, hence minimum}) \]

### Step 6: Conclusion
From our analysis, we conclude:
- \( x = 1 \) is a local maximum.
- \( x = 0 \) and \( x = 3 \) are local minima.

### Summary of Results
- Local Maximum at \( x = 1 \)
- Local Minima at \( x = 0 \) and \( x = 3 \)