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Lösung Definitions- und Wertemengen

Zuletzt geändert von Henrik Horstmann am 2026/10/19 21:33

  1. \(\begin{aligned} &D_f = \mathbb{R}^* \quad \text{und} \quad W_f = \mathbb{R}^* \\ &D_g = \mathbb{R} \quad \text{und} \quad W_g = [4,\infty[ \\ &f(g(x)) = \frac{1}{x^2+4} \\ &g(x) \neq 0 \iff x^2+4 \neq 0 \iff x^2 \neq -4 \implies x \in \mathbb{R} \\ &D_{f \circ g} = \mathbb{R} \quad \text{und} \quad W_{f \circ g} = \left]0,\frac{1}{4}\right] \end{aligned}\)

  2. \(\begin{aligned} &D_f = \mathbb{R}_+ \quad \text{und} \quad W_f = \mathbb{R}_+ \\ &D_g = \mathbb{R} \quad \text{und} \quad W_g = \mathbb{R} \\ &f(g(x)) = \sqrt{2x - 6} \\ &g(x) \geq 0 \iff 2x - 6 \geq 0 \iff 2x \geq 6 \iff x \geq 3 \\ &D_{f \circ g} = [3,\infty[ \quad \text{und} \quad W_{f \circ g} = [0,\infty[ \end{aligned}\)

  3. \(\begin{aligned} &D_f = \mathbb{R} \setminus \{2\} \quad \text{und} \quad W_f = \mathbb{R}^* \\ &D_g = \mathbb{R} \quad \text{und} \quad W_g = [-1,\infty[ \\ &f(g(x)) = \frac{1}{x^2-3} \\ &g(x) = 2 \iff x^2-1 = 2 \iff x^2 = 3 \iff x = \pm\sqrt{3} \\ &D_{f \circ g} = \mathbb{R} \setminus \{-\sqrt{3},\sqrt{3}\} \\ &\qquad W_{f \circ g} = \left]-\infty,-\frac{1}{3}\right[ \cup \left]0,\infty\right[ \end{aligned}\)