Consider the function `f(x) = 0.75 x^(4)- x^(3) - 9x^(2) +7`
What is the maximum value of the function?

Options

A. 1
B. 3
C. 7
D. 9

Answer

To find the maximum value of the function \( f(x) = 0.75 x^4 - x^3 - 9x^2 + 7 \), we will follow these steps:

### Step 1: Find the first derivative \( f'(x) \)
We start by differentiating the function to find the critical points.

\[ f'(x) = \frac{d}{dx}(0.75 x^4 - x^3 - 9x^2 + 7) \]
Using the power rule for differentiation, we get:

\[ f'(x) = 4 \cdot 0.75 x^3 - 3x^2 - 18x \]
\[ f'(x) = 3x^3 - 3x^2 - 18x \]

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

\[ 3x^3 - 3x^2 - 18x = 0 \]

### Step 3: Factor the equation
We can factor out \( 3x \):

\[ 3x(x^2 - x - 6) = 0 \]

Now, we solve for \( x \):

1. \( 3x = 0 \) gives \( x = 0 \)
2. For the quadratic \( x^2 - x - 6 = 0 \), we can factor it as:

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

This gives us the solutions:

- \( x = 3 \)
- \( x = -2 \)

So, the critical points are \( x = 0, -2, 3 \).

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

\[ f''(x) = \frac{d}{dx}(3x^3 - 3x^2 - 18x) \]
Using the power rule again, we get:

\[ f''(x) = 9x^2 - 6x - 18 \]

### Step 5: Evaluate the second derivative at the critical points
We will evaluate \( f''(x) \) at the critical points to determine the nature of each point.

1. **At \( x = 0 \)**:
\[ f''(0) = 9(0)^2 - 6(0) - 18 = -18 < 0 \quad (\text{Local maximum}) \]

2. **At \( x = -2 \)**:
\[ f''(-2) = 9(-2)^2 - 6(-2) - 18 = 36 + 12 - 18 = 30 > 0 \quad (\text{Local minimum}) \]

3. **At \( x = 3 \)**:
\[ f''(3) = 9(3)^2 - 6(3) - 18 = 81 - 18 - 18 = 45 > 0 \quad (\text{Local minimum}) \]

### Step 6: Find the maximum value
Since \( x = 0 \) is a local maximum, we find the value of the function at this point:

\[ f(0) = 0.75(0)^4 - (0)^3 - 9(0)^2 + 7 = 7 \]

### Conclusion
The maximum value of the function \( f(x) \) is:

\[ \boxed{7} \]