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

Options

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

Explanation

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

### Step 1: Find the first derivative \( f'(x) \)
We start by differentiating the function with respect to \( x \):

\[ f'(x) = \frac{d}{dx}(0.75x^4 - x^3 - 9x^2 + 7) \]

Using the power rule, we get:

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

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

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

Factoring out \( 3x \):

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

This gives us:

\[ x = 0 \quad \text{or} \quad x^2 - x - 6 = 0 \]

### Step 3: Solve the quadratic equation
Now we solve the quadratic equation \( x^2 - x - 6 = 0 \) using the factorization method:

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

Thus, the solutions are:

\[ x = 3 \quad \text{and} \quad x = -2 \]

### Step 4: Find the second derivative \( f''(x) \)
Next, we find the second derivative to determine the nature of the critical points:

\[ f''(x) = \frac{d}{dx}(3x^3 - 3x^2 - 18x) = 9x^2 - 6x - 18 \]

### Step 5: Evaluate the second derivative at critical points
Now we evaluate \( f''(x) \) at the critical points \( x = 0, 3, -2 \):

1. For \( x = 0 \):

\[ f''(0) = 9(0)^2 - 6(0) - 18 = -18 \quad (\text{concave down, local maximum}) \]

2. For \( x = 3 \):

\[ f''(3) = 9(3)^2 - 6(3) - 18 = 81 - 18 - 18 = 45 \quad (\text{concave up, local minimum}) \]

3. For \( x = -2 \):

\[ f''(-2) = 9(-2)^2 - 6(-2) - 18 = 36 + 12 - 18 = 30 \quad (\text{concave up, local minimum}) \]

### Step 6: Evaluate the function at critical points
Now, we evaluate \( f(x) \) at the critical points to find the maximum value:

1. \( f(0) \):

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

2. \( f(3) \):

\[ f(3) = 0.75(3)^4 - (3)^3 - 9(3)^2 + 7 \]
\[ = 0.75(81) - 27 - 81 + 7 = 60.75 - 27 - 81 + 7 = -0.25 \]

3. \( f(-2) \):

\[ f(-2) = 0.75(-2)^4 - (-2)^3 - 9(-2)^2 + 7 \]
\[ = 0.75(16) + 8 - 36 + 7 = 12 + 8 - 36 + 7 = -9 \]

### Conclusion
The maximum value of the function occurs at \( x = 0 \):

\[ \text{Maximum value} = f(0) = 7 \]