Lassen sich die Schnittpunkte der Graphen zweier Funktionen nicht einfach berechnen, so muß auf eine Approximation zurückgegriffen werden. Auf dieser Seite ist das Verfahren der Intervallhalbierung dargestellt. Dabei wird hier jeweils mit einer Intervallbreite von 1/10 in der Nähe der Schnittpunkte begonnen. In den Lösungen sind jeweils die ersten zehn Zeilen der entsprechenden Intervallschachtelung für die x-Koordinate des Schnittpunktes (x_S) angegeben.

Aufgaben:

Teil	Funktionsgleichung 1	Funktionsgleichung 2	Startintervall(e)
a)	y = x4	y = x + 1	[ - 0,8 ; - 0,7 ] und [ 1,2 ; 1,3 ]
b)	y = 2x3 - 5x	y = -x2 - 3	[ - 2,1 ; - 2,0 ]
c)	y = x3 - 2x2	y = 2x2 - x + 3	[ 3,9 ; 4,0 ]
d)	y = - 1/2x + 4	y = - x3 - x	[ - 1,5 ; - 1,4 ]
e)	y = 1/4x3 + 1/2 x2- 1	y = 1/3x3	[ - 1,3 ; - 1,2 ] , [ 1,6 ; 1,7 ] und [ 5,6 ; 5,7 ]

Lösung zu a)

Bedingung: x⁴ - x - 1 = 0

Erster Schnittpunkt:

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	-0,8	< xS <	-0,7
2	-0,75	< xS <	-0,7
3	-0,725	< xS <	-0,7
4	-0,725	< xS <	-0,7125
5	-0,725	< xS <	-0,71875
6	-0,725	< xS <	-0,721875
7	-0,725	< xS <	-0,7234375
8	-0,725	< xS <	-0,72421875
9	-0,724609375	< xS <	-0,72421875
10	-0,724609375	< xS <	-0,724414063

Zweiter Schnittpunkt:

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	1,2	< xS <	1,3
2	1,2	< xS <	1,25
3	1,2	< xS <	1,225
4	1,2125	< xS <	1,225
5	1,21875	< xS <	1,225
6	1,21875	< xS <	1,221875
7	1,2203125	< xS <	1,221875
8	1,2203125	< xS <	1,22109375
9	1,220703125	< xS <	1,22109375
10	1,220703125	< xS <	1,220898438

Lösung zu b)

Bedingung: 2x³ + x² - 5x + 3 = 0

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	-2,1	< xS <	-2,0
2	-2,1	< xS <	-2,05
3	-2,075	< xS <	-2,05
4	-2,075	< xS <	-2,0625
5	-2,06875	< xS <	-2,0625
6	-2,065625	< xS <	-2,0625
7	-2,0640625	< xS <	-2,0625
8	-2,0640625	< xS <	-2,06328125
9	-2,063671875	< xS <	-2,06328125
10	-2,063671875	< xS <	-2,063476563

Lösung zu c)

Bedingung: x³ - 4x² + x - 3 = 0

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	3,9	< xS <	4
2	3,9	< xS <	3,95
3	3,925	< xS <	3,95
4	3,9375	< xS <	3,95
5	3,9375	< xS <	3,94375
6	3,9375	< xS <	3,940625
7	3,9390625	< xS <	3,940625
8	3,9390625	< xS <	3,93984375
9	3,939453125	< xS <	3,93984375
10	3,939453125	< xS <	3,939648438

Lösung zu d)

Bedingung: x³ + 1/2x + 4 = 0

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	-1,5	< xS <	-1,4
2	-1,5	< xS <	-1,45
3	-1,5	< xS <	-1,475
4	-1,4875	< xS <	-1,475
5	-1,4875	< xS <	-1,48125
6	-1,484375	< xS <	-1,48125
7	-1,4828125	< xS <	-1,48125
8	-1,4828125	< xS <	-1,48203125
9	-1,4828125	< xS <	-1,482421875
10	-1,482617188	< xS <	-1,482421875

Lösung zu e)

Bedingung: - x³/12 + x²/2 - 1 = 0

Erster Schnittpunkt:

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	-1,3	< xS <	-1,2
2	-1,3	< xS <	-1,25
3	-1,3	< xS <	-1,275
4	-1,2875	< xS <	-1,275
5	-1,2875	< xS <	-1,28125
6	-1,284375	< xS <	-1,28125
7	-1,284375	< xS <	-1,2828125
8	-1,28359375	< xS <	-1,2828125
9	-1,28359375	< xS <	-1,283203125
10	-1,28359375	< xS <	-1,283398438

Zweiter Schnittpunkt:

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	1,6	< xS <	1,7
2	1,65	< xS <	1,7
3	1,65	< xS <	1,675
4	1,6625	< xS <	1,675
5	1,6625	< xS <	1,66875
6	1,6625	< xS <	1,665625
7	1,6625	< xS <	1,6640625
8	1,66328125	< xS <	1,6640625
9	1,66328125	< xS <	1,663671875
10	1,663476563	< xS <	1,663671875

Dritter Schnittpunkt:

Zeile	linke Intervallgrenze		rechte Intervallgrenze
1	5,6	< xS <	5,7
2	5,6	< xS <	5,65
3	5,6	< xS <	5,625
4	5,6125	< xS <	5,625
5	5,61875	< xS <	5,625
6	5,61875	< xS <	5,621875
7	5,61875	< xS <	5,6203125
8	5,61953125	< xS <	5,6203125
9	5,619921875	< xS <	5,6203125
10	5,619921875	< xS <	5,620117188