The function `x^(5)-5x^(4)+ 5x^(3) -10` has a maxima, when `x=`

Options

A. 3
B. 2
C. 1
D. 0

Explanation

To find the value of \( x \) at which the function \( f(x) = x^5 - 5x^4 + 5x^3 - 10 \) has a maximum, we will follow these steps:

### Step 1: Find the first derivative
We start by differentiating the function with respect to \( x \):

\[ f'(x) = \frac{d}{dx}(x^5 - 5x^4 + 5x^3 - 10) \]

Using the power rule for differentiation, 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:

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

We can factor out \( 5x^2 \):

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

### Step 3: Solve for \( x \)
Setting each factor to zero gives us:

1. \( 5x^2 = 0 \) which implies \( x = 0 \)
2. \( x^2 - 4x + 3 = 0 \)

To solve the quadratic equation \( x^2 - 4x + 3 = 0 \), we can factor it:

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

Thus, we have:

\[ x = 1 \quad \text{and} \quad x = 3 \]

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

### Step 4: Find the second derivative
Next, we need to determine whether these critical points correspond to a maximum or minimum by finding the second derivative:

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

Calculating the second derivative, we have:

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

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

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

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

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

### Conclusion
The function \( f(x) = x^5 - 5x^4 + 5x^3 - 10 \) has a maximum at:

\[ \boxed{1} \]