Hs9 Exponentiële en logaritmische functies

Diagnostische Toets Wis B VWO Hoofdstuk 9 Exponentiele en logaritmische functies Getal en Ruimte wiskunde uitwerkingen www.uitwerkingensite.nl
9.5
Diagnostische toets
bladzijde 42
1
a 3log(5) + 2 ⋅ 3log(2) = 3log(5) + 3log(22) = 3log(5 ⋅ 22) = 3log(20)
b 3 - 2log(5) = 2log(23) - 2log(5) = 2log(8) - 2log(5) = 2 log( 85 )
c
2
1
log(8000) + 3 ⋅ 2 log( 15 ) = 2log(8000) + 2 log(( 15)3 ) = 2 log(8000) - 2 log( 125
) = 2 log( 8000
) = 2log(64) = 6
125
2
a 2 ⋅ 2log(x - 1) = 1 + 2log(18)
2
log((x - 1)2) = 2log(2) + 2log(18)
2
log(x2 - 2x + 1) = 2log(36)
x2 - 2x + 1 = 36
x2 - 2x - 35 = 0
(x - 7)(x + 5) = 0
x = 7 ∨ x = -5
voldoet voldoet niet
b 2log(x) = 3 - 2log(x + 2)
2
log(x) = 2log(23) - 2log(x + 2)
 8 
log(x) = 2 log 
 x + 2
2
x=
8
x+2
x2 + 2x = 8
x2 + 2x - 8 = 0
(x + 4)(x - 2) = 0
x = - 4 ∨ x = 2
voldoet niet voldoet
3
1
a 2log(x) - 2 log(x - 1) = 3
2
log(x) + 2log(x - 1) = 3
2
log(x2 - x) = 3
x2 - x = 8
x2 - x - 8 = 0
D = 1 - 4 ⋅ 1 ⋅ -8 = 33
1+ 33
1 − 33
∨ x=
2
2
voldoet niet voldoet
b log2(x) - 5 ⋅ log(x) = 6
Stel log(x) = p.
p2 - 5p = 6
p2 - 5p - 6 = 0
(p - 6)(p + 1) = 0
p = 6 ∨ p = -1
log(x) = 6 ∨ log(x) = -1
x = 106 ∨ x = 10-1
1
x = 1 000 000 ∨ x = 10
x=
voldoet
4
34Hoofdstuk9
voldoet
a 3 + 6 ⋅( ) = 5
1
3x + 6 ⋅ x = 5
3
Stel 3x = p.
1
p+6⋅ =5
p
p2 + 6 = 5p
p2 - 5p + 6 = 0
(p - 2)(p - 3) = 0
p = 2 ∨ p = 3
3x = 2 ∨ 3x = 3
x = 3log(2) ∨ x = 1
voldoet
voldoet
x
1 x
3
b 9x = 3x + 12
(3x)2 = 3x + 12
Stel 3x = p.
p2 = p + 12
p2 - p - 12 = 0
(p - 4)(p + 3) = 0
p = 4 ∨ p = -3
3x = 4 ∨ 3x = -3
x = 3log(4) geen opl.
c 9x = 3x + 1 + 4
(3x)2 = 3x ⋅ 3 + 4
Stel 3x = p.
p2 = 3p + 4
p2 - 3p - 4 = 0
(p - 4)(p + 1) = 0
p = 4 ∨ p = -1
3x = 4 ∨ 3x = -1
x = 3log(4) geen opl.
d 3x + 2 + 32x + 1
= 12
3x ⋅ 32 + 32x ⋅ 3 = 12
9 ⋅ 3x + 3 ⋅ 32x = 12
Stel 3x = p.
9p + 3p2 = 12
p2 + 3p - 4 = 0
(p + 4)(p - 1) = 0
p = -4 ∨ p = 1
3x = - 4 ∨ 3x = 1
geen opl.
x = 0
5
a y = 3x
verm. t.o.v. de x-as met 13
↓y =
1
3
⋅ 3x
Er geldt y = 13 ⋅ 3x = 3-1 ⋅ 3x = 3x - 1.
Dusdetranslatie(1,0)levertdezelfdebeeldfiguurop.
b y = 3log(x)
↓ translatie (0, -2)
y = 3log(x) - 2
 x
Er geldt y = 3log(x) - 2 = 3log(x) - 3log(32) = 3 log   = 3log( 91 x).
 9
Dus de vermenigvuldiging ten opzichte van de y-asmet9levertdezelfdebeeldfiguurop.
6
a f (p) - g(p) = 2
3p - 1 - 4 - (2 - 3p) = 2
3p - 1 - 4 - 2 + 3p = 2
3p ⋅ 3-1 + 3p = 8
1
⋅ 3p + 3p = 8
3
4
3
⋅ 3p = 8
3p = 6
p = 3log(6)
b f (p) = g(p + 1) = q
3p - 1 - 4 = 2 - 3p + 1
3p ⋅ 3-1 - 4 = 2 - 3p ⋅ 3
1
3
⋅ 3p - 4 = 2 - 3 ⋅ 3p
∨
∨
∨
∨
∨
g(p) - f (p) = 2
2 - 3p - (3p - 1 - 4) = 2
2 - 3p - 3p - 1 + 4 = 2
-3p - 3p ⋅ 3-1 = - 4
3p + 13 ⋅ 3p = 4
∨ 43 ⋅ 3p = 4
∨
∨
∨
∨
∨
3p = 3
p=1
g(p) = f (p + 1) = q
2 - 3p = 3p + 1 - 1 - 4
2 - 3p = 3p - 4
∨ -2 ⋅ 3p = - 6
3 13 ⋅ 3p = 6
∨ 3p = 3
3p = 1 54
∨ p=1
p = 3 log( 95 )
∨ p=1
9
5
9
log( 5 ) - 1
3
p = log( ) geeft q = f (p) = 3
3
3
-4= 3
9
log( 5 )
⋅ 3-1 - 4 = 95 ⋅ 13 - 4 = 53 - 4 = -3 52
p = 1 geeft q = g(p) = 2 - 31 = 2 - 3 = -1
Exponentiëleenlogaritmischefuncties35
7
y
a
g
A
B
C
y=q
ƒ
x
O
x = –3
Stel xB = p dus xC = 3p.
f (3p) = g(p) = q geeft 2log(3p + 3) = 2log(4p)
3p + 3 = 4p
-p = -3
p=3
q = g(p) = g(3) = 2log(4 ⋅ 3) = 2log(12)
y
b
g
ƒ
F
E
D
O
x = –3
x
x=p
E is het midden van DF, dus f (p) = 2 ⋅ g(p).
2
log(p + 3) = 2 ⋅ 2log(4p)
2
log(p + 3) = 2log((4p)2)
p + 3 = (4p)2
p + 3 = 16p2
16p2 - p - 3 = 0
D = 1 - 4 ⋅ 16 ⋅ -3 = 193
1+ 193
1 − 193
≈ 0,47 ∨ p =
32
32
voldoet niet
voldoet
p=
8
a
2e3 - e3 e3
= 2 =e
e2
e
b (e3x - 5)2 = (e3x)2 - 2 ⋅ 5 ⋅ e3x + 52 = e6x - 10 e3x + 25
9
36Hoofdstuk9
a 3x ex - ex = 0
(3x - 1)ex = 0
3x - 1 = 0 ∨ ex = 0
3x = 1
geen opl.
x = 13
b e2x - 1 - 3 e 2 = 0
e2x - 1 = 3 e 2
2
e2x - 1 = e 3
2x - 1 = 23
2x = 1 23
x = 65
c e4x - ex + 1 = 0
e4x = e
x + 1
4x = x + 1
3x = 1
x = 13
d e2x + 2 ex = 3
(ex)2 + 2 ex = 3
Stel ex = p.
p2 + 2p = 3
p2 + 2p - 3 = 0
(p + 3)(p - 1) = 0
p = -3 ∨ p = 1
ex = -3 ∨ ex = 1
geen opl. x = 0
bladzijde 43
10 a f (x) = 2 ex - 3x2 geeft f ′(x) = 2 ex - 6x
b f (x) =
x 2 + 1
e x ⋅ 2x - (x 2 + 1) ⋅ e x (2x - x 2 - 1)e x - x 2 + 2x - 1
geeft
f
′(x)
=
=
=
ex
(e x )2
e2 x
ex
c f (x) = (x2 + 1)ex geeft f ′(x) = 2x ⋅ ex + (x2 + 1) ⋅ ex = (x2 + 2x + 1)ex
d f (x) =
e x
(x 2 + 1) ⋅ e x - e x ⋅ 2x (x 2 - 2x + 1)e x
geeft f ′(x) =
=
x + 1
(x 2 + 1)2
(x 2 + 1)2
2
e f (x) = x2 ⋅ e2x - 1 geeft f ′(x) = 2x ⋅ e2x - 1 + x2 ⋅ 2 ⋅ e2x - 1 = (2x2 + 2x)e2x - 1
f f (x) = e x
11
a f (x) =
2
+9
= eu met u = x2 + 9 geeft f ′(x) = eu ⋅ 2x = e x
2
+9
⋅ 2x = 2x e x
2
+9
e x
x ⋅ e x - e x ⋅1 (x - 1)e x
geeft f ′(x) =
=
x
x2
x2
f ′(x) = 0 geeft
(x - 1)e x
=0
x2
x
(x - 1)e = 0
x - 1 = 0 ∨ ex = 0
x=1
geen opl.
y
ƒ
O
x
1
min. is f (1) = e
b Stel l: y = ax + b met a = f ′(2) =
l: y = 14 e 2 x + b
e 2
f (2) = = 12 e 2 , dus A(2, 12 e 2 )
2
Dus l: y = 14 e 2 x.
}
(2 - 1) ⋅ e 2 1 2
= 4e .
22
e = 14 e 2 ⋅ 2 + b
0=b
1 2
2
Exponentiëleenlogaritmischefuncties37
1
1
12 a ln(e3 ⋅ e ) = ln(e3 ⋅ e 2 ) = ln(e 2 ) = 3 1
2
3
 1
b ln  2  = ln(e-2) = -2
e 
13 a 4 + ln(3) = ln(e4) + ln(3) = ln(3e4)
b ln(10) - 4ln(2) = ln(10) - ln(24) = ln(10) - ln(16) = ln( 10
) = ln( 85 )
16
14 a 2 ln(5x) = 16
ln(5x) = 8
5x = e8
x = 15 e8
b ln2(5x) = 16
ln(5x) = 4 ∨ ln(5x) = - 4
5x = e4 ∨ 5x = e- 4
1
x = 15 e 4 ∨ x = 15 e - 4 = 4
5e
c 2 ln2(x) - ln(x) = 0
Stel ln(x) = p.
2p2 - p = 0
p(2p - 1) = 0
p = 0 ∨ 2p - 1 = 0
p = 0 ∨ p = 12
ln(x) = 0 ∨ ln(x) = 12
1
x = e0 = 1 ∨ x = e 2
= e
d ln(9x + 1) - ln(x + 2) = ln(4)
ln(9x + 1) = ln(x + 2) + ln(4)
ln(9x + 1) = ln(4(x + 2))
ln(9x + 1) = ln(4x + 8)
9x + 1 = 4x + 8
5x = 7
x = 75 = 1 52
voldoet
15 a f (x) = 23x - 4 = 2u met u = 3x - 4 geeft f ′(x) = 2u ⋅ ln(2) ⋅ 3 = 23x - 4 ⋅ ln(2) ⋅ 3 = 3 ⋅ 23x - 1 ⋅ ln(2)
b f (x) = x ⋅ 3x geeft f ′(x) = 1 ⋅ 3x + x ⋅ 3x ⋅ ln(3) = (1 + x ln(3)) ⋅ 3x
1
1 4
1
c f (x) = ln(x ⋅ 3 x ) = ln(x 3
) = 1 13 ln( x ) geeft f ′(x) = 1 13 ⋅ =
x 3x
1
2
2
2
d f (x) = log(4x) = log(4) + log(x) geeft f ′(x) =
x ln( 2)
1
5
3
3
e f (x) = log(5x - 6) = log(u) met u = 5x - 6 geeft f ′(x) =
⋅5=
u ln(3)
(5x - 6)ln(3)
1
6x
f f (x) = ln(3x2 + 3) = ln(u) met u = 3x2 + 3 geeft f ′(x) = ⋅ 6x = 2
u
3x + 3
16 a f (x) = 3x - 1 + 3-x + 1 geeft f ′(x) = 3x - 1 ⋅ ln(3) + -1 ⋅ 3-x + 1 ⋅ ln(3) = 3x - 1 ⋅ ln(3) - 3-x + 1 ⋅ ln(3)
f ′(x) = 0 geeft 3x - 1 ⋅ ln(3) - 3-x + 1 ⋅ ln(3) = 0
3x -
1 ⋅ ln(3) = 3-x + 1 ⋅ ln(3)
3x - 1 = 3-x + 1
x - 1 = -x + 1
2x = 2
x = 1
y
ƒ
O
1
x
f (1) = 30 + 30 = 1 + 1 = 2
Bf = [2, →〉
38Hoofdstuk9
b f ′(x) = 83 ln(3) geeft 3x - 1 ⋅ ln(3) - 3-x + 1 ⋅ ln(3) = 83 ln(3)
3x - 1 - 3-x + 1 = 83
3x ⋅ 3-1 - 3-x ⋅ 31 = 83
⋅ 3x - 3 ⋅ 3-x = 83
3
1
⋅ 3x - x = 83
3
3
Stel 3x = p.
3 8
1
⋅p- = 3
3
p
1
3
1
3
p 2 - 3 = 83 p
p2 - 8p - 9 = 0
(p - 9)(p + 1) = 0
p = 9 ∨ p = -1
3x = 9 ∨ 3x = -1
x = 2 geen opl.
f (2) = 32 - 1 + 3-2 + 1 = 3 + 3-1 = 3 13
Het raakpunt is ( 2, 3 13 ).
ln( x )
=0
x
ln(x) = 0
x = 1
Het snijpunt met de x-as is (1, 0).
1
x ⋅ - ln( x ) ⋅1 1 - ln( x )
ln( x )
=
f (x) =
geeft f ′(x) = x 2
x
x
x2
1 - ln(1)
Stel k: y = ax + b met a = f ′(1) =
= 1.
12
17 a f (x) = 0 geeft
k: y = x + b
door (1, 0)
}
0=1+b
-1 = b
Dus k: y = x - 1.
1 - ln( x )
b f ′(x) = 0 geeft
=0
x2
1 - ln(x) = 0
ln(x) = 1
x=e
y
ƒ
O
e
x
ln(e ) 1
=
e
e
1
Bf =  , →
e
f (e) =
Exponentiëleenlogaritmischefuncties39

Download Report