Let f(x) = `40 (3x^(4) + 8x^(3) - 18x^(2) + 60)` , consider the following statement about f(x) .

Options

A. f(x) has local minima at x = 0
B. f(x) has local maxima at x = 0
C. absolute maximum value of f(x) is not defined
D. f(x) is local maxima at x = –3, x = 1

Answer

To solve the problem, we need to analyze the function \( f(x) = 40(3x^4 + 8x^3 - 18x^2 + 60) \) for its maxima and minima. Here’s a step-by-step solution:

### Step 1: Differentiate the function
First, we find the first derivative \( f'(x) \) of the function \( f(x) \).

\[ f(x) = 40(3x^4 + 8x^3 - 18x^2 + 60) \]

Using the power rule for differentiation:

\[ f'(x) = 40(12x^3 + 24x^2 - 36x) \]

### Step 2: Set the first derivative to zero
To find the critical points, we set \( f'(x) = 0 \):

\[ 40(12x^3 + 24x^2 - 36x) = 0 \]

Dividing both sides by 40:

\[ 12x^3 + 24x^2 - 36x = 0 \]

Factoring out \( 12x \):

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

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

1. \( 12x = 0 \) → \( x = 0 \)
2. \( x^2 + 2x - 3 = 0 \)

Factoring the quadratic:

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

Thus, we have:

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

The critical points are \( x = 0, -3, 1 \).

### Step 4: Find the second derivative
Next, we find the second derivative \( f''(x) \):

\[ f'(x) = 40(12x^3 + 24x^2 - 36x) \]

Differentiating again:

\[ f''(x) = 40(36x^2 + 48x - 36) \]

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

1. **At \( x = 0 \)**:

\[ f''(0) = 40(36(0)^2 + 48(0) - 36) = 40(-36) < 0 \quad \text{(Local Maxima)} \]

2. **At \( x = -3 \)**:

\[ f''(-3) = 40(36(-3)^2 + 48(-3) - 36) = 40(324 - 144 - 36) = 40(144) > 0 \quad \text{(Local Minima)} \]

3. **At \( x = 1 \)**:

\[ f''(1) = 40(36(1)^2 + 48(1) - 36) = 40(36 + 48 - 36) = 40(48) > 0 \quad \text{(Local Minima)} \]

### Step 6: Conclusion
- Local maxima at \( x = 0 \)
- Local minima at \( x = -3 \) and \( x = 1 \)