If f(x)={(-x^3/2)+x+2, x>-1
{ 1-(x/2) x
Range of f(x) where x-[-2,-1] union (1,2) is NOT contained in
(A) (0,5/2) (B) (-1,3) (c) [3/2,15/2] (d) (0,3)

To determine the range of the function \( f(x) \) defined piecewise, we need to analyze both parts of the function separately and then combine the results. The function is given as follows:

For \( x > -1 \): \( f(x) = -\frac{x^{3}}{2} + x + 2 \)

For \( x \in [-2, -1] \): \( f(x) = 1 - \frac{x}{2} \)

Step 1: Analyzing the first piece of the function

Let's first look at the function for \( x > -1 \). The expression \( -\frac{x^{3}}{2} + x + 2 \) is a cubic polynomial, which typically has a range that can be determined by finding its critical points and evaluating its behavior as \( x \) approaches infinity.

Finding critical points

To find the critical points, we need to take the derivative of \( f(x) \) and set it to zero:

1. Calculate the derivative:

\( f'(x) = -\frac{3x^{2}}{2} + 1 \)

2. Set the derivative equal to zero:

\( -\frac{3x^{2}}{2} + 1 = 0 \)

\( \frac{3x^{2}}{2} = 1 \)

\( x^{2} = \frac{2}{3} \)

\( x = \pm \sqrt{\frac{2}{3}} \) (only \( x = \sqrt{\frac{2}{3}} \) is relevant since we are considering \( x > -1 \))

Evaluating the function at critical points and endpoints

Next, we evaluate \( f(x) \) at \( x = \sqrt{\frac{2}{3}} \) and also consider the limit as \( x \) approaches infinity:

1. Calculate \( f(\sqrt{\frac{2}{3}}) \):

\( f(\sqrt{\frac{2}{3}}) = -\frac{(\sqrt{\frac{2}{3}})^{3}}{2} + \sqrt{\frac{2}{3}} + 2 \)

2. As \( x \to \infty \), \( f(x) \to -\infty \) (since the cubic term dominates).

Step 2: Analyzing the second piece of the function

Now, let's examine the function for \( x \in [-2, -1] \): \( f(x) = 1 - \frac{x}{2} \).

This is a linear function, and we can find its range by evaluating it at the endpoints of the interval:

1. At \( x = -2 \): \( f(-2) = 1 - \frac{-2}{2} = 1 + 1 = 2 \)

2. At \( x = -1 \): \( f(-1) = 1 - \frac{-1}{2} = 1 + 0.5 = 1.5 \)

Thus, for \( x \in [-2, -1] \), the range of \( f(x) \) is \( [1.5, 2] \).

Combining the ranges

Now, we need to combine the ranges from both parts of the function:

• From \( x \in [-2, -1] \): Range is \( [1.5, 2] \).
• From \( x > -1 \): The function decreases to \( -\infty \) and has a local maximum at \( f(\sqrt{\frac{2}{3}}) \) which we need to calculate for completeness.

Finding the overall range

Since the cubic function will eventually go to \( -\infty \) and has a local maximum, we can conclude that the overall range of \( f(x) \) is \( (-\infty, 2] \).

Identifying the excluded intervals

Now, we need to determine which of the given options does not fall within this range:

• (A) \( (0, \frac{5}{2}) \) - Partially included.
• (B) \( (-1, 3) \) - Partially included.
• (C) \( [\frac{3}{2}, \frac{15}{2}] \) - Partially included.
• (D) \( (0, 3) \) - Partially included.

Since the range of \( f(x) \) is \( (-\infty, 2] \), the interval that is NOT contained in the range is the one that exceeds 2. Therefore, the answer is:

None of the intervals exceed 2, so all options are partially included. However, if we consider the maximum value of the cubic function, we find that the range is limited to values less than or equal to 2.