Lösungen Test 1a

Umkehrfunktionen,Definitionsbereich etc.

1

a

f[x_] := Tan[x];  g[x_] :=  Cos[x];  
    h[x_] := Sin[x];
f1[x_] := ArcTan[x];  g1[x_] :=  ArcCos[x];  
    h1[x_] := ArcSin[x];

p1 = Plot[ArcSin[x], {x, -1, 1}];

p2 = Plot[Floor[ArcSin[x]], {x, -1, 1}];

Show[p1, p2];

h[g[x]]

{g[x], h[y], g[h[x]]}

b

g[x]=y,  h[y]=z, y=h1[z], x=g1[y]=g1[h1[z]]

{h1[z], g1[y], g1[h1[z]], g1[h1[x]]}

hgInv[x_] := ArcCos[ArcSin[x]]

p3 = Plot[{ArcSin[x], hgInv[x]}, {x, -1, 1}];

(===> Mathematica meckert wegen dem Definitionsbereich!)

FindRoot[ArcSin[x] == 1, {x, 0.5}]

FindRoot[ArcSin[x] == -1, {x, -0.5}]

c

p4 = Plot[h[g[f[x]]], {x, -Pi, Pi}];

p4 = Plot[h[g[f[x]]], {x, -Pi/2, Pi/2}];

Ungleichungen

Solve[x^2 - 2  x - 3 ==  x + 2, x]

% // N

Solve[x^2 - 2  x - 3 ==  -3 x^2 + x + 7, x]

% // N

Solve[ x + 2 == -3 x^2 + x + 7, x]

% // N

Plot[{x^2 - 2  x - 3,  x + 2, -3 x^2 + x + 7}, {x, -10, 
          10}];

Plot[{x^2 - 2  x - 3,  
          x + 2, -3 x^2 + x + 7}, {x, -1.4, -1}];

Plot[{x^2 - 2  x - 3,  x + 2, -3 x^2 + x + 7}, {x, 1, 
          3}];

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}  /.  x -> -5

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}  /.  x -> -1.28

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> -1.22

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /. { x ->  (3 - Sqrt[29])/2}

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /. { 
        x ->  (3 - Sqrt[29])/2 - 0.000001}

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /. { 
        x ->  (3 - Sqrt[29])/2 + 0.000001}

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> 0

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> 1.5

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> 2

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> 2 + 0.000001

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> 2.5

{x^2 - 2  x - 3 <=  
          x + 2, 
x^2 - 2  x - 3 <=  -3 x^2 + x + 
            7}   /.  x -> 5

Lösung

Intervall [(1/2) (3 - (Sqrt[29])), 
        2] approximiert durch [-1.1925824035672519, 2]

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> -5

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> -1.28

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> -1.22

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> 0

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> 1.5

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> 2.5

{x^2 - 2  x - 3 >  
          x + 2, 
x^2 - 2  x - 3 >  -3 x^2 + x + 
            7, 
 x + 2 > -3 x^2 + x + 7}  /.  
      x -> 5

Pole, Asymptoten,...

3. Asymtote, Pole, Nullstellen, Definitionsbereich, ungefährer Wertebereich

Remove[x]

a[x_] := (x^3 - 2  x)/(x^2 - 4)  + 
          1;
a[x]

Division:

Apart[a[x]]

a) Pole:

Solve[x^2 - 4 == 0, {x}]

b) Asymptote:

Aus  1 + ((-2 x + (x^3))/(-4 + (x^2))) =  1 + 
            (1/(-2 + x)) + x + 
            (1/(2 + x))     === > y = 
        x + 1

1-Stellen:

Solve[a[x] == 1, {x}]

e) Plot:

Plot[{x + 1, a[x]}, {x, -9, 10}];

Definitionsbereich

Klar! (Siehe Graph!),  x  ungleich +/- 2 (an den Polstellen nicht definiert!)