The number of real roots of the equation `2x^(3) -3x^(2) + 6x + 6 = 0` is

Options

A. 1
B. 2
C. 3
D. none of these

Answer

To determine the number of real roots of the equation \(2x^3 - 3x^2 + 6x + 6 = 0\), we can follow these steps:

### Step 1: Define the function
Let \(f(x) = 2x^3 - 3x^2 + 6x + 6\).

### Step 2: Find the first derivative
To analyze the behavior of the function, we first find its first derivative:
\[ f'(x) = \frac{d}{dx}(2x^3 - 3x^2 + 6x + 6) = 6x^2 - 6x + 6 \]

### Step 3: Analyze the first derivative
Next, we will analyze the first derivative to determine where it is positive or negative:
\[ f'(x) = 6(x^2 - x + 1) \]
Now, we need to check if \(x^2 - x + 1 = 0\) has any real roots. The discriminant of this quadratic is:
\[ D = b^2 - 4ac = (-1)^2 - 4(1)(1) = 1 - 4 = -3 \]
Since the discriminant is negative, \(x^2 - x + 1\) has no real roots and is always positive. Thus, \(f'(x) > 0\) for all \(x\).

### Step 4: Conclusion from the first derivative
Since \(f'(x) > 0\) for all \(x\), the function \(f(x)\) is strictly increasing. This means that it can cross the x-axis at most once.

### Step 5: Evaluate the function at a specific point
To check if there is indeed a root, we can evaluate \(f(x)\) at a couple of points:
- At \(x = 0\):
\[ f(0) = 2(0)^3 - 3(0)^2 + 6(0) + 6 = 6 \]
- At \(x = -1\):
\[ f(-1) = 2(-1)^3 - 3(-1)^2 + 6(-1) + 6 = -2 - 3 - 6 + 6 = -5 \]

### Step 6: Apply the Intermediate Value Theorem
Since \(f(-1) < 0\) and \(f(0) > 0\), by the Intermediate Value Theorem, there is at least one root in the interval \((-1, 0)\).

### Final Conclusion
Since \(f(x)\) is strictly increasing and crosses the x-axis only once, we conclude that there is exactly **one real root** of the equation \(2x^3 - 3x^2 + 6x + 6 = 0\).

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