Pembahasan 15 Soal Invers Fungsi Komposisi



 1.) Diketahui f(x) = 8x – 3a, dengan a  0. Jika f-1(5) = 4, maka nilai √a adalah . . .

A. 3
B. -3
C. 4
D. -5
E 5

Pembahasan :
f(x) = 8x – 3a
y = 8x – 3a
8x = y + 3a
x = (y + 3a)/8
f-1(x) = (x + 3a)/8

Maka,
f-1(x) = (x + 3a)/8
f-1(5) = (5 + 3a)/8
4 = (5 + 3a)/8
32 = 5 + 3a
3a = 27
a = 9

Jadi,
= √a
= √9
= 3 (A)

2.) Jika g(x + 5) = (2x – 1)/(x + 3), maka g-1(-2) = . . .

A. 15/4
B. 5
C. 15/2
D. 15
E. 3

Pembahasan :

g(x + 5) = (2x – 1)/(x + 3)
misal : a = x + 5 → x = a – 5

g(a) = (2(a – 5) – 1)/(a – 5 + 3)
g(a) = (2a – 11)/(a – 2)

Maka,
g(x) = (2x – 11)/(x – 2)

Buat invers :
y = (2x – 11)/(x – 2)
y(x – 2) = 2x – 11
xy – 2y = 2x – 11
xy – 2x = 2y – 11
x(y – 2) = 2y – 11
x = (2y – 11)/(y – 2)
g-1(x) = (2x – 11)/(x – 2)
g-1(-2) = (2(-2) – 11)/(-2 – 2)
g-1(-2) = -15/-4
g-1(-2) = 15/4 (A)

3.) Diketahui f(x) = 2x – 7, invers dari f(4x + 3) = . . .

A. 2x – 5
B. 5 – 2x
C. 5x – 2
D. 2 – 5x
E. 2x + 5

Pembahasan :
f(x) = 2x – 7
y = 2x – 7
2x = y + 7
x = (y + 7)/2
f-1(x) = (x + 7)/2

Maka,
f-1(4x + 3) = (4x + 3 + 7)/2
f-1(4x + 3) = (4x + 10)/2
f-1(4x + 3) = 2x + 5 (E)

4.) Diketahui f(x) = (4x – 2)/(x – 1) dengan f-1(x) merupakan invers dari f(x). Jika f(x) = f-1(x), maka persamaannya adalah . . .

A. x2 – 5x – 2 = 0
B. x2 + 5x + 2 = 0
C. x2 – 5x + 2 = 0
D. 2x2 – 5x + 2 = 0
E. 2x2 + 5x – 2 = 0

Pembahasan :
f(x) = (3x – 2)/(x – 1)
y = (3x – 2)/(x – 1)
y(x – 1) = 4x – 2
xy – y = 4x – 2
xy – 4x = y – 2
x(y – 4) = y – 2
x = (y – 2)/(y – 4)
f-1(x) = (x – 2)/(x – 4)

Maka,
f(x) = f-1(x)
(4x – 2)/(x – 1) = (x – 2)/(x – 4)
(4x – 2)(x – 4) = (x – 2)(x – 1)
4x2 – 18x + 8 = x2 – 3x + 2
4x2 – x2 – 18x + 3x + 8 – 2 = 0
3x2 – 15x + 6 = 0
x2 – 5x + 2 = 0 (C)

5.) Diketahui f(x) = 2x – 6 dan g(x) = 5 – x. Invers dari f-1(g-1(-1)) = . . .

A. 6
B. 8
C. 5
D. 2
E. 3

Pembahasan :
f(x) = 2x – 6
y = 2x – 6
2x = y + 6
x = (y + 6)/2
f-1(x) = (x + 6)/2

g(x) = 5 – x
y = 5 – x
x = 5 – y
g-1(x) = 5 – x

Maka,
f-1(g-1(x)) = 2(5 – x) – 6
f-1(g-1(x)) = 10 – 2x – 6
f-1(g-1(x)) = 4 – 2x
f-1(g-1(-1)) = 4 – 2(-1)
f-1(g-1(-1)) = 4 + 2
f-1(g-1(-1)) = 6 (A)

6.) Invers dari f(x) = x2 – 6 adalah . . .

A. x + 6
B. √(x – 6)
C. x – 6
D. √(x + 6)
E. (x + 6)2

Pembahasan :
f(x) = x2 – 6
y = x2 – 6
x2 = y + 6
x = √(y + 6)
f-1(x) = √(x + 6) (D)

7.) Diketahui fungsi p(x) = 9x – 2 dan q(x) = (2x – 1)/x . Invers dari (q o p)-1(1) = . . .

A. 1/3
B. 1/4
C. 1/5
D. 1/6
E. 1/7

Pembahasan :
p(x) = 9x – 2
y = 9x – 2
9x = y + 2
x = (y + 2)/9
p-1(x) = (x + 2)/9

q(x) = (2x – 1)/x
y = (2x – 1)/x
xy = 2x – 1
xy – 2x = -1
x(y – 2) = -1
x = -1/(y – 2)
q-1(x) = -1/(x – 2)

Maka,
(q o p)-1(x) = p-1(q-1(x))
= [(-1/(x – 2) + 2)]/9

p-1(q-1(1)) = [(-1/(1 – 2) + 2)]/9
= (1 + 2)/9
= 3/9
= 1/3 (A)

8.) Diketahui fungsi f(x) = x2 + 6x + 9. Invers dari f(9) = . . .

A. 4
B. 3
C. 2
D. 1
E. 0

Pembahasan :
f(x) = x2 + 6x + 9
y = x2 + 6x + 9
y = (x + 3)2
√y = x + 3
x = √y – 3
f-1(x) = √x – 3

Maka,
f-1(9) = √9 – 3
f-1(9) = 3 – 3
f-1(9) = 0 (E)

9.) Jika fungsi f(x) = √3x – 6. Maka invers dari f-1(3) = . . .

A. 25
B. 26
C. 27
D. 28
E. 29

Pembahasan :
f(x) = √3x – 6
y = √3x – 6
√3x = y + 6
3x = (y + 6)2
3x = y2 + 12y + 36
x = (y2 + 12y + 36)/3
f-1(x) = (x2 + 12x + 36)/3

Maka,
f-1(3) = ((3)2 + 12(3) + 36)/3
= (9 + 36 + 36)/3
= 3 + 12 + 12
= 27 (C)

10.) Diketahui f(2x – 1) = 1 – 2x dan g(x +1) = x – 4. Invers (f o g)-1(x) = . . .

A. -x + 5
B. x – 5
C. -x – 5
D. x + 5
E. 2x – 5

Pembahasan :

f(2x – 1) = 1 – 2x
misal,
a = 2x – 1
x = (a + 1)/2

f(a) = 1 – 2((a + 1)/2)
f(a) = 1 – a + 1
f(a) = -a
f(x) = -x

f(x) = -x
f-1(x) = -x

g(x + 1) = x – 4
misal,
a = x + 1
x = a – 1

g(a) = a – 1 – 4
g(a) = a – 5
g(x) = x – 5

g(x) = x – 5
y = x – 5
x = y + 5
g-1(x) = x + 5

Maka,
(f o g)-1(x) = (g-1 o f-1)(x)
(f o g)-1(x) = (-x) + 5
(f o g)-1(x) = -x + 5 (A)

11.) Diketahui fungsi f(x) = √(2x – 3). Invers f(-3) = . . .

A. 3
B. 4
C. 5
D. 6
E. 7

Pembahasan :
f(x) = √(2x – 3)
y = √(2x – 3)
y2 = 2x – 3
2x = y2 + 3
x = (y2 + 3)/2

Maka,
f-1(x) = (x2 + 3)/2
f-1(-3) = ((-3)2 + 3)/2
f-1(-3) = (9 + 3)/2
f-1(-3) = 12/2
f-1(-3) = 6 (D)

12.) Diketahui f(x) = -2x + 3. Invers f(x) adalah . . .

A. (3 + x)/2
B. (x – 3)/2
C. (3 – x)/3
D. (3 + x)/3
E. (3 – x)/2

Pembahasan :
f(x) = -2x + 3
y = -2x + 3
2x = 3 – y
x = (3 – y)/2

Jadi,
f-1(x) = (3 – x)/2 (E)

13.) Diketahui g(x + 1) = (x – 1)/(x + 1). Invers g(0) = . . .

A. 4
B. -4
C. -2
D. 1
E. 2

Pembahasan :

g(x + 1) = (x – 1)/(x + 1)
misal,
a = x + 1
x = a – 1

g(a) = (a – 2)/a
g(x) = (x – 2)/x

Maka,
g(x) = (x – 2)/x
y = (x – 2)/x
xy = x – 2
x – xy = 2
x(1 – y) = 2
x = 2/(1 – y)
g-1(x) = 2/(1 – x)
g-1(0) = 2/(1 – 0)
g-1(0) = 2 (E)

14.) Diberikan f(x) = x + a. Jika f(x) = f-1(x), maka nilai a adalah . . .

A. 2
B. 1
C. 0
D. -1
E. -2

Pembahasan :
f(x) = x + a
y = x + a
x = y – a
f-1(x) = x – a

Maka,
f(x) = f-1(x)
x + a = x – a
a + a = x – x
2a = 0
a = 0 (C)

15.) Diberikan fungsi g(x) = √2x + √6. Jika g-1(x) merupakan invers dari g(x), maka g-1(2√6) = . . .

A. 2
B. 3
C. 5
D. 6
E. 8

Pembahasan :
g(x) = √2x + √6
y = √2x + √6
√2x = y – √6
2x = (y – √6)2
x = (y – √6)2/2

Maka,
g-1(x) = (x – √6)2/2
g-1(2√6) = (2√6 – √6)2/2
g-1(2√6) = (√6)2/2
g-1(2√6) = 6/2
g-1(2√6) = 3 (B)

Post a Comment

Previous Post Next Post

Contact Form