Soal/Latihan Fungsi dan Komposisi (Pertemuan ke-10)
Soal
1. Diketahui π(π₯) = √π₯ dan π(π₯) = π₯ + 1, maka selesaikan :
a) π(π₯) ∘ π(π₯)
b) π(π₯) ∘ π(π₯)
c) π(π₯) ∘ π(π₯)
d) π(π₯) ∘ π(π₯)
2. Selesaikan soal berikut,
π(π₯) = π₯^2 + 2π₯ + 5 dan π(π₯) = 3π₯ maka carilah
a) π(π₯) ∘ π(π₯)
b) π(π₯) ∘ π(π₯)
c) π(π₯) + π(π₯)
d) π(π₯) − π(π₯)
3. Diketahui f(x) = x2 + 1 dan g(x) = 2x − 3, maka (f o g)(x) = ….
Diketahui fungsi f(x) = 3x − 1 dan g(x) = π₯^2 + 2π₯ + 5 . Nilai dari komposisi fungsi
(π ∘ π)(1) dan (π ∘ π)(2)
4. Diberikan dua buah fungsi:
f(x) = 2x − 3
g(x) = x2 + 2x + 3
Jika (f o g)(a) = 33, tentukan nilai dari 5a
Jawab
1.
a. (π ∘ π)(π₯) = (π(π(π₯)) =
b. (π ∘ π)(π₯) = (π(π(π₯)) = √π₯ + 1
c. (π ∘ π)(π₯) = (π(π(π₯)) = √√π₯ =
d. (π ∘ π)(π₯) = (π(π(π₯)) = (π₯ + 1) + 1 = π₯ + 2
2.
a. π(π₯) ∘ π(π₯) = (π(π(π₯)) = π(3x) = 9x^2 +6x+5
b. π(π₯) ∘ π(π₯) = (π ∘ π)(π₯) = π(π₯^2 + 2π₯ + 5) = 3x^2 + 2x + 5
c. π(π₯) + π(π₯) = π₯^2 + 2π₯ + 5 + 3x = π₯^2 + 5π₯ + 5
d. π(π₯) − π(π₯) = 3x - π₯^2 + 2π₯ + 5 = -x^2 + 5x + 5
3.
a. (π ∘ π)(π₯) = (π(π(π₯)) = π(2x - 3) = (2x - 3)2 + 1 = 4x - 5
b.
(π ∘ π)(π₯) = (π(π(π₯)) = π(x^2 + 2x +5) = 3(x^2 + 2x + 5) - 1 = 3x^2 + 6x + 29
(π ∘ π)(π₯) = (π(π(π₯)) = π(3x -1) = (3x - 1)^2 + 2(3x - 1) + 5 = 9x^ + 1
4.
(f o g)(x) = 2(x^2 + 2x + 3) - 3
= 2x^2 + 4x + 6 - 3 = 2x^2 + 4x + 3
(f o g)(a) = 33
2a^2 + 4a + 3 = 33
2a^2 + 4a - 30 = 0
(2a + 10)(a - 3)
a^2 + 2a − 15 = 0 Faktorkan:
(a + 5)(a − 3) = 0
a = -5 = 5(-5) = -25
a = 3 = 5(3) = 15
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