1. Tentukan
nilai suku banyak x2 – 7x +
10 jika x = 4!
Jawab : (4)2
– 7(4) + 10 = 16 – 28 + 10
=
-2
2. Jika f(x) = (2x-4)2 – x2, hitunglah f(2), f(0), dan f(-2)!
Jawab : f(2) =
(2(2) – 4)2 – (2)2
= (4 – 4)2 – (2)2
= -4
f(0) = (2(0) – 4)2 –
(0)2
= (-4)2
= 16
f(-2 )= (2(-2) – 4)2 – (-2)2
= (-8)2 – 4
= 64 – 4
= 60
3. f(x) = x4 + 3x3 – px2 + (p+2)x + 3 dibagi (x+2) menghasilkan sisa 15. Tentukan nilai p!
3. f(x) = x4 + 3x3 – px2 + (p+2)x + 3 dibagi (x+2) menghasilkan sisa 15. Tentukan nilai p!
Jawab : x = -2
f(-2) = 16 – 24 – 4p – 2p – 4 +
3
15 = -9 – 6p
6p = -24
P = -4
4. Diketahui g(x) = x + 3, (fog)(x) =x2 + 4, Tentukan f(x)!
Jawab : y = x + 3
x = y – 3
(fog)(x)
= x2 + 4
f(x+3) = x2 + 4
f(y) = (y – 3)2 + 4
f(y) = y2 – 6y + 9 + 4
f(y) = y2 – 6y + 13
f(x) = x2 – 6x + 13
5. Jika (f o g)(x) = 4x2 + 8x - 3 dan g(x) = 2x + 4, maka tentukan f-1(x)!
5. Jika (f o g)(x) = 4x2 + 8x - 3 dan g(x) = 2x + 4, maka tentukan f-1(x)!
Jawab
: f(g(x)) = 4x2 + 8x - 3
f(2x + 4) = 4x2 + 8x - 3
f(x)
= x2 - 4x -
3 berarti : a = 1, b = -4, dan c = -3
f-1(x) = {-b ± √(b2 - 4a(c -x)}/ 2a
f-1(x) =
{4 ± √(16- 4(-3 -x)}/ 2
f-1(x) = {4 ± √(16 + 12 + 4x)}/ 2
f-1(x) = {4 ± √(28 + 4x)}/ 2
f-1(x) =
{4 ± √(4(7 + x))}/ 2
f-1(x) =
{4 ± 2√(7 + x)}/ 2
f-1(x) =
2 ± √(7 + x)
6. Hitunglah lim x2 – 1 !
x → -1 x + 1
Jawab : (x
+ 1) (x – 1) = x – 1
= -1 -1
= -2
7. Hitunglah lim sin (x – 1)
7. Hitunglah lim sin (x – 1)
x → 1 x2
-1
Jawab : lim
sin (x – 1) = lim sin
(x – 1)
x → 1 x2 -1 x → 1 (x
– 1) (x + 1)
=
lim 1/ x + 1
x → 1
= 1 / 1 + 1
= 1/2
8. Hitunglah f(x) = sin x . cos x
8. Hitunglah f(x) = sin x . cos x
Jawab : f(x)
= sin x . cos x
= (sin x) (-sin x) + (cos x) (cos x)
= - sin2x + cos2x
= -(1 – cos2x) + cos2x
= 2cos2x - 1
= cos 2x
9. Hitunglah f(x) = (4x2 + 5x) (2x2 – 6x + 1)
Jawab :
f(x) = (4x2 + 5x) (2x2
– 6x + 1)
= u’
. v + u . v’
= (8x
+ 5) (2x2 – 6x + 1) + (4x2 + 5x) (4x – 6)
=
(16x3 – 48x2 + 8x + 10x2 – 30x + 5) + (16x3
– 24x2 + 20x2 – 30x)
= 32x3
– 42x2 – 52x + 5
10. Tentukan nilai maksimum mutlak dan minimum mutlak pada interval tertutup yang diketahui
f(x) = 4x2
– 4x + 1, pada interval [ 0,1]!
Jawab : f(x)
= 4x2 – 4x + 1
y’ = 8x – 4
x = ½
f(1/2) = 4(1/2)2 –
4(1/2) + 1
= 1 – 2 +1 = 0
f(0) = 4(0)2 – 4(0) + 1
= 0 – 0 + 1 = 1
f(1) = 4(1)2 – 4(1) + 1
= 4 – 4 + 1 = 1
Jadi nilai
maksimum adalah 1, dan nilai minimum adalah 0
2 Comments:
makasih ilmunya
makasih kak
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