The function $f(x) = |x^2 + x - 2| - 4$ involves the absolute value of a quadratic function. Here's a step-by-step analysis of how to study this function:

Step 1: Analyze the quadratic inside the absolute value

The expression inside the absolute value is: $x^2 + x - 2$ Factorize it: $x^2 + x - 2 = (x - 1)(x + 2)$

Step 2: Determine where the quadratic changes sign

The roots of $x^2 + x - 2$ are $x = 1$ and $x = -2$. These roots divide the number line into three intervals:

• $(-\infty, -2)$
• $(-2, 1)$
• $(1, \infty)$

Check the sign of $x^2 + x - 2$ in each interval:

• For $x \in (-\infty, -2)$: Both factors $(x - 1)$ and $(x + 2)$ are negative, so the product is positive.
• For $x \in (-2, 1)$: The factor $(x - 1)$ is negative, and $(x + 2)$ is positive, so the product is negative.
• For $x \in (1, \infty)$: Both factors are positive, so the product is positive.

Thus:

\begin{cases} x^2 + x - 2, & \text{if } x \in (-\infty, -2) \cup (1, \infty), \\ -(x^2 + x - 2), & \text{if } x \in (-2, 1). \end{cases}$$ ### Step 3: Write $$f(x)$$ without the absolute value Using the analysis above: $$f(x) = \begin{cases} (x^2 + x - 2) - 4, & \text{if } x \in (-\infty, -2) \cup (1, \infty), \\ -(x^2 + x - 2) - 4, & \text{if } x \in (-2, 1). \end{cases}$$ Simplify each case: 1. For $$x \in (-\infty, -2) \cup (1, \infty)$$: $$f(x) = x^2 + x - 6$$ 2. For $$x \in (-2, 1)$$: $$f(x) = -x^2 - x - 2$$ ### Step 4: Analyze critical points and behavior 1. **Critical points**: The critical points occur where $$f'(x) = 0$$ or where $$f(x)$$ is not differentiable. - The derivative of $$f(x) = x^2 + x - 6$$ is $$f'(x) = 2x + 1$$. - The derivative of $$f(x) = -x^2 - x - 2$$ is $$f'(x) = -2x - 1$$. Check for non-differentiable points at $$x = -2$$ and $$x = 1$$ (where the absolute value changes): - At $$x = -2$$: $$f(x) = 6$$. - At $$x = 1$$: $$f(x) = -5$$. 2. **Domain**: $$f(x)$$ is defined for all real $$x$$. 3. **Range**: To determine the range, analyze the vertex of each quadratic piece. - For $$x \in (-\infty, -2) \cup (1, \infty)$$: The vertex of $$x^2 + x - 6$$ is at $$x = -\frac{1}{2}$$, and the parabola opens upwards. - For $$x \in (-2, 1)$$: The vertex of $$-x^2 - x - 2$$ is at $$x = -\frac{1}{2}$$, and the parabola opens downwards. ### Step 5: Graphical representation The function is piecewise quadratic, with a smooth transition at $$x = -2$$ and $$x = 1$$. The absolute value causes $$f(x)$$ to change behavior at these points. --- Would you like a detailed plot of this function or further numerical analysis? ### Relative Questions: 1. What are the roots of $$f(x) = 0$$? 2. How does the function behave at the critical points? 3. What is the range of $$f(x)$$? 4. How does the absolute value affect the symmetry of the function? 5. Can we use this function to model any physical phenomena? **Tip**: When working with absolute values in functions, always break the analysis into intervals where the expression inside the absolute value is positive and negative.