The number of points of maxima/minima of `f(x) =x(x + 1) (x +2) (x + 3)` is

Options

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

Answer

To find the number of points of maxima and minima of the function \( f(x) = x(x + 1)(x + 2)(x + 3) \), we will follow these steps:

### Step 1: Rewrite the function
First, we can expand the function to make it easier to differentiate:
\[ f(x) = x(x + 1)(x + 2)(x + 3) \]
We can group and multiply:
\[ f(x) = (x(x + 3))((x + 1)(x + 2)) = (x^2 + 3x)(x^2 + 3x + 2) \]

### Step 2: Differentiate the function
Now, we will differentiate \( f(x) \) using the product rule. Let:
\[ u = x^2 + 3x \quad \text{and} \quad v = x^2 + 3x + 2 \]
Then, the derivative \( f'(x) \) is given by:
\[ f'(x) = u'v + uv' \]
Calculating the derivatives:
\[ u' = 2x + 3 \quad \text{and} \quad v' = 2x + 3 \]
Thus,
\[ f'(x) = (2x + 3)(x^2 + 3x + 2) + (x^2 + 3x)(2x + 3) \]
Factoring out \( (2x + 3) \):
\[ f'(x) = (2x + 3)(x^2 + 3x + 2 + x^2 + 3x) = (2x + 3)(2x^2 + 6x + 2) \]

### Step 3: Set the derivative to zero
To find the critical points, we set the derivative equal to zero:
\[ (2x + 3)(2x^2 + 6x + 2) = 0 \]
This gives us two equations to solve:
1. \( 2x + 3 = 0 \)
2. \( 2x^2 + 6x + 2 = 0 \)

### Step 4: Solve for critical points
From the first equation:
\[ 2x + 3 = 0 \implies x = -\frac{3}{2} \]

For the second equation, we can simplify:
\[ 2x^2 + 6x + 2 = 0 \implies x^2 + 3x + 1 = 0 \]
Using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-3 \pm \sqrt{5}}{2} \]
This gives us two more critical points:
\[ x_1 = \frac{-3 + \sqrt{5}}{2}, \quad x_2 = \frac{-3 - \sqrt{5}}{2} \]

### Step 5: Count the critical points
We have found three critical points:
1. \( x = -\frac{3}{2} \)
2. \( x = \frac{-3 + \sqrt{5}}{2} \)
3. \( x = \frac{-3 - \sqrt{5}}{2} \)

### Conclusion
The total number of points of maxima and minima of the function \( f(x) = x(x + 1)(x + 2)(x + 3) \) is **3**.

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