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Lösung Verkettungen bilden

Zuletzt geändert von Anna Kukin am 2026/07/24 18:28

  1. \(u(x) = 2+x\) ; \(v(x) = 3x-2\)
    \(f(x) = u(3x-2) = 2 + (3x-2) = 3x\)
    \(g(x) = v(2+x) = 3(2+x) - 2 = 6 + 3x - 2 = 3x + 4\)

  2. \(u(x) = \frac{1}{x}\) ; \(v(x) = x+3\)
    \(f(x) = u(x+3) = \frac{1}{x+3}\)
    \(g(x) = v\left(\frac{1}{x}\right) = \frac{1}{x} + 3\)
  3. \(u(x) = x^2-2\) ; \(v(x) = (x-2)^2\)
    \(f(x) = u\left((x-2)^2\right) = \left((x-2)^2\right)^2 - 2 = (x-2)^4 - 2\)
    \(g(x) = v(x^2-2) = \left((x^2-2) - 2\right)^2 = (x^2-4)^2\)
  4. \(u(x) = e^x\) ; \(v(x) = 1-x\)
    \(f(x) = u(1-x) = e^{1-x}\)
    \(g(x) = v(e^x) = 1 - e^x\)
  5. \(u(x) = \cos(x)\) ; \(v(x) = 2x+1\)
    \(f(x) = u(2x+1) = \cos(2x+1)\)
    \(g(x) = v(\cos(x)) = 2\cos(x) + 1\)
  6. \(u(x) = \frac{3}{x}\) ; \(v(x) = \frac{3}{3-x^2}\)
    \(f(x) = u\left(\frac{3}{3-x^2}\right) = \frac{3}{\frac{3}{3-x^2}} = 3 \cdot \frac{3-x^2}{3} = 3-x^2\)
    \(g(x) = v\left(\frac{3}{x}\right) = \frac{3}{3 - \left(\frac{3}{x}\right)^2} = \frac{3}{3 - \frac{9}{x^2}} = \frac{3}{\frac{3x^2 - 9}{x^2}} = \frac{3x^2}{3x^2 - 9} = \frac{x^2}{x^2 - 3}\)