`|x^(2) -1|+|x^(2) - 4|le 3`

Answer

To solve the inequality \( |x^2 - 1| + |x^2 - 4| \leq 3 \), we will analyze the expression by considering the critical points where the expressions inside the absolute values change sign. The critical points are found by setting the expressions inside the absolute values to zero:

1. \( x^2 - 1 = 0 \) gives \( x = -1, 1 \)
2. \( x^2 - 4 = 0 \) gives \( x = -2, 2 \)

Thus, the critical points are \( -2, -1, 1, 2 \). We will consider the intervals defined by these points: \( (-\infty, -2) \), \( [-2, -1) \), \( [-1, 1) \), \( [1, 2) \), and \( [2, \infty) \).

### Step 1: Analyze the interval \( (-\infty, -2) \)

In this interval, both expressions are positive:
\[ |x^2 - 1| = x^2 - 1 \quad \text{and} \quad |x^2 - 4| = x^2 - 4 \]
Thus, the inequality becomes:
\[ (x^2 - 1) + (x^2 - 4) \leq 3 \]
\[ 2x^2 - 5 \leq 3 \]
\[ 2x^2 \leq 8 \quad \Rightarrow \quad x^2 \leq 4 \]
This implies:
\[ -2 \leq x \leq 2 \]
However, since we are in the interval \( (-\infty, -2) \), there are no solutions in this interval.

### Step 2: Analyze the interval \( [-2, -1) \)

In this interval:
\[ |x^2 - 1| = x^2 - 1 \quad \text{and} \quad |x^2 - 4| = 4 - x^2 \]
The inequality becomes:
\[ (x^2 - 1) + (4 - x^2) \leq 3 \]
\[ 3 \leq 3 \]
This is always true for \( x \in [-2, -1) \).

### Step 3: Analyze the interval \( [-1, 1) \)

In this interval:
\[ |x^2 - 1| = 1 - x^2 \quad \text{and} \quad |x^2 - 4| = 4 - x^2 \]
The inequality becomes:
\[ (1 - x^2) + (4 - x^2) \leq 3 \]
\[ 5 - 2x^2 \leq 3 \]
\[ -2x^2 \leq -2 \quad \Rightarrow \quad x^2 \geq 1 \]
This implies:
\[ x \leq -1 \quad \text{or} \quad x \geq 1 \]
However, since we are in the interval \( [-1, 1) \), the only solution is \( x = -1 \).

### Step 4: Analyze the interval \( [1, 2) \)

In this interval:
\[ |x^2 - 1| = x^2 - 1 \quad \text{and} \quad |x^2 - 4| = 4 - x^2 \]
The inequality becomes:
\[ (x^2 - 1) + (4 - x^2) \leq 3 \]
\[ 3 \leq 3 \]
This is always true for \( x \in [1, 2) \).

### Step 5: Analyze the interval \( [2, \infty) \)

In this interval:
\[ |x^2 - 1| = x^2 - 1 \quad \text{and} \quad |x^2 - 4| = x^2 - 4 \]
The inequality becomes:
\[ (x^2 - 1) + (x^2 - 4) \leq 3 \]
\[ 2x^2 - 5 \leq 3 \]
\[ 2x^2 \leq 8 \quad \Rightarrow \quad x^2 \leq 4 \]
This implies:
\[ -2 \leq x \leq 2 \]
However, since we are in the interval \( [2, \infty) \), the only solution is \( x = 2 \).

### Final Solution

Combining all the intervals, we have:
\[ [-2, -1) \cup \{-1\} \cup [1, 2) \]
Thus, the final solution set is:
\[ [-2, -1] \cup [1, 2] \]