1. **Problem statement:** Given functions $f(x) = x^2 - 4$ and $g(x) = x + 2$, find the composite function $(f \circ g)(x)$, determine its domain and range, and investigate its monotony. 2. **Formula and rules:** The composite function $(f \circ g)(x)$ means $f(g(x))$. To find it, substitute $g(x)$ into $f(x)$. 3. **Find $(f \circ g)(x)$:** $$ (f \circ g)(x) = f(g(x)) = f(x + 2) = (x + 2)^2 - 4 $$ 4. **Expand and simplify:** $$ (x + 2)^2 - 4 = (x^2 + 4x + 4) - 4 = x^2 + 4x $$ 5. **Domain:** Since $g(x) = x + 2$ is defined for all real $x$, and $f$ is defined for all real numbers, the domain of $(f \circ g)(x)$ is all real numbers, i.e., $\mathbb{R}$. 6. **Range:** The function $(f \circ g)(x) = x^2 + 4x$ is a quadratic. Complete the square to find its range: $$ x^2 + 4x = (x^2 + 4x + 4) - 4 = (x + 2)^2 - 4 $$ Since $(x + 2)^2 \geq 0$, the minimum value is $-4$ at $x = -2$. Thus, the range is $[-4, \infty)$. 7. **Monotony:** To investigate monotony, find the derivative: $$ \frac{d}{dx} (f \circ g)(x) = \frac{d}{dx} (x^2 + 4x) = 2x + 4 $$ - For $2x + 4 > 0$, i.e., $x > -2$, the function is increasing. - For $2x + 4 < 0$, i.e., $x < -2$, the function is decreasing. - At $x = -2$, the derivative is zero, indicating a minimum point. **Final answers:** - $(f \circ g)(x) = x^2 + 4x$ - Domain: $\mathbb{R}$ - Range: $[-4, \infty)$ - Monotony: decreasing on $(-\infty, -2)$, increasing on $(-2, \infty)$ with a minimum at $x = -2$.