Wiki-Quellcode von Lösung Verkettungen bilden
Zuletzt geändert von Anna Kukin am 2026/07/24 18:28
Zeige letzte Bearbeiter
| author | version | line-number | content |
|---|---|---|---|
| 1 | (%class=abc%) | ||
| 2 | 1. ((({{formula}}u(x) = 2+x{{/formula}} ; {{formula}}v(x) = 3x-2{{/formula}} | ||
| 3 | {{formula}}f(x) = u(3x-2) = 2 + (3x-2) = 3x{{/formula}} | ||
| 4 | {{formula}}g(x) = v(2+x) = 3(2+x) - 2 = 6 + 3x - 2 = 3x + 4{{/formula}}))) | ||
| 5 | 1. {{formula}}u(x) = \frac{1}{x}{{/formula}} ; {{formula}}v(x) = x+3{{/formula}} | ||
| 6 | {{formula}}f(x) = u(x+3) = \frac{1}{x+3}{{/formula}} | ||
| 7 | {{formula}}g(x) = v\left(\frac{1}{x}\right) = \frac{1}{x} + 3{{/formula}} | ||
| 8 | 1. {{formula}}u(x) = x^2-2{{/formula}} ; {{formula}}v(x) = (x-2)^2{{/formula}} | ||
| 9 | {{formula}}f(x) = u\left((x-2)^2\right) = \left((x-2)^2\right)^2 - 2 = (x-2)^4 - 2{{/formula}} | ||
| 10 | {{formula}}g(x) = v(x^2-2) = \left((x^2-2) - 2\right)^2 = (x^2-4)^2{{/formula}} | ||
| 11 | 1. {{formula}}u(x) = e^x{{/formula}} ; {{formula}}v(x) = 1-x{{/formula}} | ||
| 12 | {{formula}}f(x) = u(1-x) = e^{1-x}{{/formula}} | ||
| 13 | {{formula}}g(x) = v(e^x) = 1 - e^x{{/formula}} | ||
| 14 | 1. {{formula}}u(x) = \cos(x){{/formula}} ; {{formula}}v(x) = 2x+1{{/formula}} | ||
| 15 | {{formula}}f(x) = u(2x+1) = \cos(2x+1){{/formula}} | ||
| 16 | {{formula}}g(x) = v(\cos(x)) = 2\cos(x) + 1{{/formula}} | ||
| 17 | 1. {{formula}}u(x) = \frac{3}{x}{{/formula}} ; {{formula}}v(x) = \frac{3}{3-x^2}{{/formula}} | ||
| 18 | {{formula}}f(x) = u\left(\frac{3}{3-x^2}\right) = \frac{3}{\frac{3}{3-x^2}} = 3 \cdot \frac{3-x^2}{3} = 3-x^2{{/formula}} | ||
| 19 | {{formula}}g(x) = v\left(\frac{3}{x}\right) = \frac{3}{3 - \left(\frac{3}{x}\right)^2} = \frac{3}{3 - \frac{9}{x^2}} = \frac{3}{\frac{3x^2 - 9}{x^2}} = \frac{3x^2}{3x^2 - 9} = \frac{x^2}{x^2 - 3}{{/formula}} |