L¨ Pr¨
Aufgabe 9.1
f:y = 7x−5.
f(0) =−5 7x−5 = 0 7x= 5 x= 57 Aufgabe 9.2
f:y =x2+ 9.
f(0) = 9 x2+ 9 = 0
x2 =−9 keine Nullstellen Aufgabe 9.3
f:y = 3x2−2x.
f(0) = 0 3x2−2x= 0 x 3x−2
= 0 x1 = 0 3x2−2 = 0 3x2 = 2 x2 = 23
f:y =x2−8.
f(0) =−8 x2−8 = 0 x2 = 8 x1 =√
8 = 2√ 2 x2 =−√
8 = 2√ 2 Aufgabe 9.5
f:y =x2−4x−1.
f(0) =−1
x2−4x−1 = 0; Koeffizienten: a= 1, b =−4, c=−1
Diskriminante: D=b2−4ac= 16−4·1·(−1) = 16 + 4 = 20 x1 = −b+√
D
2a = 4 +√ 20
2 = 4 + 2√ 5
2 = 2 +√ 5 x2 = −b−√
D
2a =· · ·= 2−√ 5
Aufgabe 9.6 f:y =x4−5x2+ 6.
f(0) = 6
x4−5x2+ 6 = 0 Substitution:x2 =u
u2 −5u+ 6 = 0 (u−2)(u−3) = 0 u1 = 2 u2 = 3
Substitution r¨uckg¨angig machen:
u1 = 2 =x2 ⇒ x1 =√
2,x2 =−√ 2 u2 = 3 =x2 ⇒ x3 =√
3,x4 =−√ 3
f:y =x11−4x9. f(0) = 0
x11−4x9 = 0 x10(x2−4) = 0 x1 = 0 x2 = 2 x3 =−2 Aufgabe 9.8
f:y = (x+ 1)(x−√ 3)2. f(0) = 1· −√
32
= 3 (x+ 1)(x−√
3)2 = 0x1 =−1 x2 =x3 =√
3 Aufgabe 9.9
f:y = (x+ 5)(x−3) x(x−4) . f(0) = 5·(−3)
0·(−4) = −15
0 nicht definiert (x+ 5)(x−3)
x(x−4) = 0 x1 =−5 x2 = 3 Aufgabe 9.10 f:y = x2−7x+ 12
x2−5x+ 6 . f(0) = 12
6 = 2 x2 −7x+ 12
x2−5x+ 6 = 0 (x−3)(x−4) (x−2)(x−3) = 0
f:y =√
x−3−2.
f(0) =√
−3−2 nicht definiert
√x−3−2 = 0
√x−3 = 2 x−3 = 4 x= 7
Probe: (in der urspr¨unglichen Gleichung)
√7−3−2 = √
4−2 = 0 x= 7
Aufgabe 9.12 f:y =√
x+ 2 + 3.
f(0) = 3 +√ 2
√x+ 2 + 3 = 0
√x+ 2 =−3 x+ 2 = 9
x= 7
Probe: (in der urspr¨unglichen Gleichung)
√7 + 2 + 3 =√
9 + 3 = 9 (falsch) keine Nullstellen
Aufgabe 9.13 f:y = ex−2.
f(0) = e0 −2 = 1−2 =−1 ex−2 = 0
ex = 2 || ln(. . .) ln ex
= ln(2) x= ln(2)
f:y = (x−5)ex.
f(0) = (0−5)·e0 =−5·1 =−5 (x−5)ex = 0
x= 5
(die Funktion y= ex hat keine Nullstellen) Aufgabe 9.15
f:y = ln(x−7).
f(0) = ln(−7) ist nicht definiert ln(x−7) = 0 ||e...
eln(x−7) = e0 x−7 = 1
x= 8 Aufgabe 9.16 f:y = ln(x+ 1)−3.
f(0) = ln(1)−3 = 0−3 =−3 ln(x+ 1)−3 = 0
ln(x+ 1) = 3 ||e...
eln(x+1) = e3 x+ 1 = e3
x= e3−1 Aufgabe 9.17
f:y = sin(2x+ 1).
f(0) = sin(1) sin(2x+ 1) = 0
Da sin(t) = 0 f¨ur t=k·π und k ∈ {0,±1,±2, . . .} 2xk+ 1 =k·π f¨urk ∈ {0,±1,±2, . . .}
xk = k·π−1
f¨ur k∈ {0,±1,±2, . . .}
f:y = 4 cos 12x−3 .
f(0) = 4 cos(−3) = 4 cos(3) . 4 cos 12x−3
= 0 || : 4 cos 12x−3
= 0 Da cos(t) = 0 f¨ur t= π
2 +k·π und k∈ {0,±1,±2, . . .}
1
2xk−3 = π2 +k·π f¨urk ∈ {0,±1,±2, . . .} xk−6 =π+ 2k·π f¨ur k∈ {0,±1,±2, . . .}
xk= 6 + (2k+ 1)π f¨ur k ∈ {0,±1,±2, . . .} Aufgabe 9.19
f:y = tan(3x+π).
f(0) = tan(π) = 0.
tan 3x+π
= 0
Da tan(t) = 0 f¨urt =k·π und k∈ {0,±1,±2, . . .} 3xk+π =k·π f¨urk ∈ {0,±1,±2, . . .}
3xk =k·π−π f¨urk ∈ {0,±1,±2, . . .} xk = (k−1)π
3 f¨ur k ∈ {0,±1,±2, . . .} Aufgabe 9.20
1. Funktion im TI-84+ darstellen (Standardkoordinatensystem)
2. 2nd // calc // 2:zero // LeftBound? −2 // Enter // RightBound? 0 // Enter //
Guess? // Enter ⇒ x1 =−0.706
3. 2nd // calc // 2:zero // LeftBound? 0 // Enter // RightBound? 2 // Enter //
Guess? // Enter ⇒ x2 = 1.20
4. 2nd // calc // 2:zero // LeftBound? 2 // Enter // RightBound? 5 // Enter //
Guess? // Enter ⇒ x3 = 3.68
1. Funktion im TI-84+ darstellen (Standardkoordinatensystem)
2. 2nd // calc // 2:zero // LeftBound? −2 // Enter // RightBound? 0 // Enter //
Guess? // Enter ⇒ x1 =−1.13
3. 2nd // calc // 2:zero // LeftBound? 1 // Enter // RightBound? 3 // Enter //
Guess? // Enter ⇒ x2 = 1.71
4. 2nd // calc // 2:zero // LeftBound? 3 // Enter // RightBound? 5 // Enter //
Guess? // Enter ⇒ x3 = 4.06