To determine the image set (range) of the function \( f(x) = (x - 4)^2 + 4 \) for the domain \(-2 < x < 4\), we need to analyze the behavior of the function within the given interval. First, let's rewrite the function: \[ f(x) = (x - 4)^2 + 4 \] The function \( (x - 4)^2 \) is a parabola that opens upwards with its vertex at \( x = 4 \). The minimum value of \( (x - 4)^2 \) is 0, which occurs when \( x = 4 \). Therefore, the minimum value of \( f(x) \) is: \[ f(4) = (4 - 4)^2 + 4 = 0 + 4 = 4 \] Next, we need to find the maximum value of \( f(x) \) within the interval \(-2 < x < 4\). Since \( (x - 4)^2 \) increases as \( x \) moves away from 4, the maximum value of \( f(x) \) will occur at the left endpoint of the interval, \( x = -2 \): \[ f(-2) = (-2 - 4)^2 + 4 = (-6)^2 + 4 = 36 + 4 = 40 \] Therefore, the image set (range) of the function \( f(x) = (x - 4)^2 + 4 \) for \(-2 < x < 4\) is: \[ 4 < f(x) < 40 \] So, the correct answer is: **D. (4, 40)**

Sure, let's break down the solution step by step to determine the image set (range) of the function $f(x) = (x - 4)^2 + 4$ for the domain $-2 < x < 4$.

1. Rewrite the Function: The given function is: $f(x) = (x - 4)^2 + 4$

2. Analyze the Function: The function $(x - 4)^2$ is a parabola that opens upwards with its vertex at $x = 4$. The minimum value of $(x - 4)^2$ is 0, which occurs when $x = 4$.

3. Find the Minimum Value of $f(x)$: Since the minimum value of $(x - 4)^2$ is 0, the minimum value of $f(x)$ is: $f(4) = (4 - 4)^2 + 4 = 0 + 4 = 4$

4. Find the Maximum Value of $f(x)$ within the Interval $-2 < x < 4$: Since $(x - 4)^2$ increases as $x$ moves away from 4, the maximum value of $f(x)$ will occur at the left endpoint of the interval, $x = -2$: $f(-2) = (-2 - 4)^2 + 4 = (-6)^2 + 4 = 36 + 4 = 40$

5. Determine the Range: Therefore, the image set (range) of the function $f(x) = (x - 4)^2 + 4$ for $-2 < x < 4$ is: $4 < f(x) < 40$

So, the correct answer is: D. (4, 40)