Saya bantu beberapa nomor ya. 22. f(x) = (2x + 1)/(x – 3), x ≠ 3 Misalkan y = f(x) maka y = (2x + 1)/(x – 3)↔ y(x – 3) = 2x + 1 ↔ xy – 3y = 2x + 1 ↔ xy – 2x = 3y + 1 ↔ x(y – 2) = 3y + 1 ↔ x = (3y + 1)/(y – 2) Kita peroleh f⁻¹(x) = (3x + 1)/(x – 2), x ≠ 2 f⁻¹(x + 1) = (3(x + 1) + 1)/(x + 1 – 2) = (3x + 3 + 1)/(x – 1) = (3x + 4)/(x – 1) dengan x ≠ 1 Jawab: B 25. f(x) = (mx – 5)/(3x + 2), x ≠ -2/3 Misalkan y = f(x) maka y = (mx – 5)/(3x + 2) ↔ y(3x + 2) = mx – 5 ↔ 3xy + 2y = mx – 5 ↔ 3xy – mx = -2y – 5 ↔ x(3y – m) = -2y – 5 ↔ x = (-2y – 5)/(3y – m) Kita peroleh f⁻¹(x) = (-2x – 5)/(3x – m), x ≠ m/3 Untuk f¹(3) = 1, maka (-2(3) – 5)/(3(3) – m) = 1 ↔ (-6 – 5)/(9 – m) = 1 ↔ -11 = 9 – m ↔ m = 9 + 11 = 20 Jadi, m = 20 Tidak ada opsi yang tepat pada pilihan jawaban.
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Saya bantu beberapa nomor ya.
22. f(x) = (2x + 1)/(x – 3), x ≠ 3
Misalkan y = f(x) maka
y = (2x + 1)/(x – 3)
↔ y(x – 3) = 2x + 1
↔ xy – 3y = 2x + 1
↔ xy – 2x = 3y + 1
↔ x(y – 2) = 3y + 1
↔ x = (3y + 1)/(y – 2)
Kita peroleh
f⁻¹(x) = (3x + 1)/(x – 2), x ≠ 2
f⁻¹(x + 1) = (3(x + 1) + 1)/(x + 1 – 2)
= (3x + 3 + 1)/(x – 1)
= (3x + 4)/(x – 1)
dengan x ≠ 1
Jawab: B
25. f(x) = (mx – 5)/(3x + 2), x ≠ -2/3
Misalkan y = f(x) maka
y = (mx – 5)/(3x + 2)
↔ y(3x + 2) = mx – 5
↔ 3xy + 2y = mx – 5
↔ 3xy – mx = -2y – 5
↔ x(3y – m) = -2y – 5
↔ x = (-2y – 5)/(3y – m)
Kita peroleh
f⁻¹(x) = (-2x – 5)/(3x – m), x ≠ m/3
Untuk f¹(3) = 1, maka
(-2(3) – 5)/(3(3) – m) = 1
↔ (-6 – 5)/(9 – m) = 1
↔ -11 = 9 – m
↔ m = 9 + 11 = 20
Jadi, m = 20
Tidak ada opsi yang tepat pada pilihan jawaban.