If `f(x)=8-x^(2)` where `-3lt x lt3`, what is the range of values of `f(x)`?

Options

A. `0ltylt7`
B. `-1leyle8`
C. `8ltylt17`
D. `-1ltyle8`

Answer

To find the range of the function \( f(x) = 8 - x^2 \) where \( -3 < x < 3 \), we will follow these steps:

### Step 1: Identify the function and its domain
The function is given as:
\[ f(x) = 8 - x^2 \]
The domain of \( x \) is \( -3 < x < 3 \).

### Step 2: Determine the critical points
Since \( f(x) \) is a quadratic function, it opens downwards (as the coefficient of \( x^2 \) is negative). The maximum value will occur at the vertex of the parabola. The vertex of the function \( f(x) = 8 - x^2 \) can be found at \( x = 0 \) (since the function is symmetric around the y-axis).

### Step 3: Calculate the maximum value
Substituting \( x = 0 \) into the function:
\[ f(0) = 8 - (0)^2 = 8 \]
Thus, the maximum value of \( f(x) \) is \( 8 \).

### Step 4: Calculate the minimum value at the endpoints of the domain
Next, we will evaluate the function at the endpoints of the domain:
1. For \( x = -3 \):
\[ f(-3) = 8 - (-3)^2 = 8 - 9 = -1 \]
2. For \( x = 3 \):
\[ f(3) = 8 - (3)^2 = 8 - 9 = -1 \]

### Step 5: Determine the range
From the calculations:
- The maximum value of \( f(x) \) is \( 8 \) (at \( x = 0 \)).
- The minimum value of \( f(x) \) is \( -1 \) (at \( x = -3 \) and \( x = 3 \)).

Since the function is defined for \( -3 < x < 3 \), the minimum value \( -1 \) is not included in the range. Therefore, the range of \( f(x) \) is:
\[ (-1, 8] \]

### Final Answer
The range of values of \( f(x) \) is:
\[ (-1, 8] \]

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