Let `f(x)=x^3+3/2x^2+3x+3`, then f(x) is

Options

A. a decreasing function
B. an increasing function
C. an odd function
D. an even function

Answer

To determine whether the function \( f(x) = x^3 + \frac{3}{2}x^2 + 3x + 3 \) is increasing or decreasing, we need to analyze its derivative \( f'(x) \).

### Step 1: Find the derivative \( f'(x) \)

The function is given as:
\[ f(x) = x^3 + \frac{3}{2}x^2 + 3x + 3 \]

To find the derivative, we differentiate each term:
\[ f'(x) = \frac{d}{dx}(x^3) + \frac{d}{dx}\left(\frac{3}{2}x^2\right) + \frac{d}{dx}(3x) + \frac{d}{dx}(3) \]

Calculating each derivative:
- The derivative of \( x^3 \) is \( 3x^2 \).
- The derivative of \( \frac{3}{2}x^2 \) is \( 3x \).
- The derivative of \( 3x \) is \( 3 \).
- The derivative of a constant \( 3 \) is \( 0 \).

Thus, we have:
\[ f'(x) = 3x^2 + 3x + 3 \]

### Step 2: Factor out the common term

We can factor out \( 3 \) from the derivative:
\[ f'(x) = 3(x^2 + x + 1) \]

### Step 3: Analyze the quadratic \( x^2 + x + 1 \)

Next, we need to determine if \( x^2 + x + 1 \) is always positive. We can do this by calculating its discriminant \( D \):
\[ D = b^2 - 4ac \]
where \( a = 1 \), \( b = 1 \), and \( c = 1 \).

Calculating the discriminant:
\[ D = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]

### Step 4: Conclusion about the quadratic

Since the discriminant \( D < 0 \), the quadratic \( x^2 + x + 1 \) has no real roots and opens upwards (as the coefficient of \( x^2 \) is positive). Therefore, \( x^2 + x + 1 > 0 \) for all \( x \).

### Step 5: Determine the sign of \( f'(x) \)

Since \( f'(x) = 3(x^2 + x + 1) \) and \( x^2 + x + 1 > 0 \), we conclude that:
\[ f'(x) > 0 \quad \text{for all } x \]

### Final Conclusion

Since the derivative \( f'(x) \) is always positive, the function \( f(x) \) is an increasing function for all values of \( x \).