1. **Problem statement:** We have four functions: $$f_1(x) = x^2 - 2,$$ $$f_2(x) = x^2 - 1,$$ $$f_3(x) = x^2,$$ $$f_4(x) = x^2 + 1.$$ We want to find their zeros (Nullstellen), i.e., values of $x$ where $f_i(x) = 0$. 2. **Formula and rules:** To find zeros of a quadratic function $f(x) = x^2 + c$, solve $$x^2 + c = 0 \implies x^2 = -c.$$ - If $-c \geq 0$, real roots exist: $x = \pm \sqrt{-c}$. - If $-c < 0$, no real roots exist. 3. **Find zeros for each function:** - For $f_1(x) = x^2 - 2$: $$x^2 - 2 = 0 \implies x^2 = 2 \implies x = \pm \sqrt{2} \approx \pm 1.414.$$ - For $f_2(x) = x^2 - 1$: $$x^2 - 1 = 0 \implies x^2 = 1 \implies x = \pm 1.$$ - For $f_3(x) = x^2$: $$x^2 = 0 \implies x = 0.$$ - For $f_4(x) = x^2 + 1$: $$x^2 + 1 = 0 \implies x^2 = -1,$$ no real roots, so no zeros. 4. **Verification by substitution:** - Check $f_1(\pm \sqrt{2}) = (\pm \sqrt{2})^2 - 2 = 2 - 2 = 0.$ - Check $f_2(\pm 1) = (\pm 1)^2 - 1 = 1 - 1 = 0.$ - Check $f_3(0) = 0^2 = 0.$ - $f_4$ has no real zeros. **Final answers:** - $f_1$ zeros: $x \approx -1.414, 1.414$ - $f_2$ zeros: $x = -1, 1$ - $f_3$ zero: $x = 0$ - $f_4$ zeros: none (enter NaN)