The function `x^(5)-5x^(4)+5x^(3)-1` is

Options

A. neither maximum nor minimum at `x=0`
B. maximum at `x=0`
C. maximum at `x=1` and minimum at `x=3`
D. minimum at `x=0`

Explanation

To determine the maximum and minimum points of the function \( f(x) = x^5 - 5x^4 + 5x^3 - 1 \), we will follow these steps:

### Step 1: Differentiate the function
We start by finding the first derivative of the function \( f(x) \).

\[ f'(x) = \frac{d}{dx}(x^5 - 5x^4 + 5x^3 - 1) \]

Using the power rule, we differentiate each term:

\[ f'(x) = 5x^4 - 20x^3 + 15x^2 \]

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

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

### Step 3: Factor the equation
We can factor out the common term \( 5x^2 \):

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

Now, we can further factor the quadratic:

\[ 5x^2(x - 1)(x - 3) = 0 \]

### Step 4: Solve for critical points
Setting each factor to zero gives us the critical points:

\[ 5x^2 = 0 \quad \Rightarrow \quad x = 0 \]
\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

So, the critical points are \( x = 0, 1, 3 \).

### Step 5: Determine maximum and minimum using the second derivative
Next, we find the second derivative of \( f(x) \):

\[ f''(x) = \frac{d}{dx}(5x^4 - 20x^3 + 15x^2) \]

Differentiating again:

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

### Step 6: Evaluate the second derivative at critical points
Now we evaluate \( f''(x) \) at each critical point:

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

2. For \( x = 1 \):
\[ f''(1) = 20(1)^3 - 60(1)^2 + 30(1) = 20 - 60 + 30 = -10 \]
(Since \( f''(1) < 0 \), there is a local maximum at \( x = 1 \))

3. For \( x = 3 \):
\[ f''(3) = 20(3)^3 - 60(3)^2 + 30(3) = 540 - 540 + 90 = 90 \]
(Since \( f''(3) > 0 \), there is a local minimum at \( x = 3 \))

### Conclusion
Thus, we have:
- A local maximum at \( x = 1 \)
- A local minimum at \( x = 3 \)