`f(x)=x^(3)-6x^(2)+9x-2` has maximum at the point

Options

A. `(-2, 3)`
B. `(3, -2)`
C. `(2, 3)`
D. `(1, 2)`

Answer

To find the maximum point of the function \( f(x) = x^3 - 6x^2 + 9x - 2 \), we will follow these steps:

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

\[ f'(x) = \frac{d}{dx}(x^3 - 6x^2 + 9x - 2) = 3x^2 - 12x + 9 \]

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

\[ 3x^2 - 12x + 9 = 0 \]

### Step 3: Simplify the equation
We can simplify this equation by dividing through by 3:

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

### Step 4: Factor the quadratic equation
Now we will factor the quadratic equation:

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

### Step 5: Solve for \( x \)
Setting each factor to zero gives us the critical points:

\[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
\[ x - 3 = 0 \quad \Rightarrow \quad x = 3 \]

### Step 6: Find the second derivative \( f''(x) \)
To determine whether these critical points are maxima or minima, we need to find the second derivative.

\[ f''(x) = \frac{d}{dx}(3x^2 - 12x + 9) = 6x - 12 \]

### Step 7: Evaluate the second derivative at the critical points
Now we will evaluate the second derivative at each critical point:

1. For \( x = 1 \):

\[ f''(1) = 6(1) - 12 = 6 - 12 = -6 \quad (\text{less than } 0) \]

2. For \( x = 3 \):

\[ f''(3) = 6(3) - 12 = 18 - 12 = 6 \quad (\text{greater than } 0) \]

### Step 8: Determine the nature of the critical points
- Since \( f''(1) < 0 \), the function has a **local maximum** at \( x = 1 \).
- Since \( f''(3) > 0 \), the function has a **local minimum** at \( x = 3 \).

### Step 9: Find the maximum value
Now we will find the maximum value of the function at \( x = 1 \):

\[ f(1) = 1^3 - 6(1^2) + 9(1) - 2 = 1 - 6 + 9 - 2 = 2 \]

### Conclusion
The function \( f(x) = x^3 - 6x^2 + 9x - 2 \) has a maximum at the point \( (1, 2) \).

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