The function `f(x)=x^(5)-5x^(4)+5x^(3)` find maximum and minimum value

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 find the maximum and minimum values of the function \( f(x) = x^5 - 5x^4 + 5x^3 \), we will follow these steps:

### Step 1: Find the first derivative
We start by differentiating the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^5 - 5x^4 + 5x^3) \]
Using the power rule, we get:
\[ f'(x) = 5x^4 - 20x^3 + 15x^2 \]

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

\[ 5x^4 - 20x^3 + 15x^2 = 0 \]

Factoring out the common term \( 5x^2 \):

\[ 5x^2(x^2 - 4x + 3) = 0 \]

This gives us:

\[ 5x^2 = 0 \quad \text{or} \quad x^2 - 4x + 3 = 0 \]

From \( 5x^2 = 0 \), we have \( x = 0 \).

Now, we solve the quadratic equation \( x^2 - 4x + 3 = 0 \):

\[ (x - 1)(x - 3) = 0 \]

Thus, we find \( x = 1 \) and \( x = 3 \).

### Step 3: Find the second derivative
Next, we find the second derivative to determine the nature of the critical points:

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

\[ f''(x) = 20x^3 - 60x^2 + 30x \]

### Step 4: Evaluate the second derivative at critical points
Now, we evaluate the second derivative at the critical points \( x = 0, 1, 3 \):

1. For \( x = 0 \):
\[ f''(0) = 20(0)^3 - 60(0)^2 + 30(0) = 0 \]

2. For \( x = 1 \):
\[ f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 \quad (\text{less than 0, local maximum}) \]

3. For \( x = 3 \):
\[ f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 20(27) - 60(9) + 90 = 540 - 540 + 90 = 90 \quad (\text{greater than 0, local minimum}) \]

### Step 5: Conclusion
From our evaluations, we conclude:
- There is a local maximum at \( x = 1 \).
- There is a local minimum at \( x = 3 \).

### Step 6: Find the maximum and minimum values
Now, we calculate the function values at these critical points:

1. For \( x = 1 \):
\[ f(1) = 1^5 - 5(1^4) + 5(1^3) = 1 - 5 + 5 = 1 \]

2. For \( x = 3 \):
\[ f(3) = 3^5 - 5(3^4) + 5(3^3) = 243 - 405 + 135 = -27 \]

### Final Answer
- Maximum value: \( f(1) = 1 \)
- Minimum value: \( f(3) = -27 \)

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