Exercise Zone : Fungsi Komposisi [2]
Table of Contents

Tipe:
No.
Diketahui f : ℝ ⟶ ℝ dan g : ℝ ⟶ ℝ denganALTERNATIF PENYELESAIAN
\begin{aligned}\left(f\circ g\right)(x)&=f\left(g(x)\right)\\&=f(3x-2)\\&=\dfrac{3x-2-4}{3x-2+3}\\y&=\dfrac{3x-6}{3x+1}\\(3x+1)y&=3x-6\\3xy+y&=3x-6\\3xy-3x&=-y-6\\x(3y-3)&=-y-6\\x&=\dfrac{-y-6}{3y-3}\\(f\circ g)^{-1}(x)&=\dfrac{-x-6}{3x-3}\end{aligned}
Jadi,\left(f\circ g\right)^{-1} (x)=\dfrac{-x-6}{3x-3} .
No.
Diketahui : f(x) = x + 4g(x) = x2 − 4
h(x) = x − 5
Tentukan (f ∘ g ∘ h)(x) ?
ALTERNATIF PENYELESAIAN
\begin{aligned}
(f\circ g\circ h)(x)&=f(g(h(x)))\\
&=f(g(x-5))\\
&=f\left((x-5)^2-4\right)\\
&=f\left(x^2-10x+25-4\right)\\
&=f\left(x^2-10x+21\right)\\
&=x^2-10x+21+4\\
&=\boxed{\boxed{x^2-10x+25}}
\end{aligned}
Jadi, (f ∘ g ∘ h)(x) = x2 − 10x + 25.
No.
Diketahui f(a) = 12a + 4 dan g(a)= 5a − 2 tentukan (g−1 ∘ f−1)(a)ALTERNATIF PENYELESAIAN
CARA 1
CARA 2
\begin{aligned} (f\circ g)(x)&=f(g(x))\\ &=f(5a-2)\\ &=12(5a-2)+4\\ &=60a-24+4\\ &=60a-20 \end{aligned} \begin{aligned} \left(g^{-1}\circ f^{-1}\right)(a)&=(f\circ g)^{-1}(a)\\ &=\boxed{\boxed{\dfrac{a+20}{60}}} \end{aligned}Jadi, \left(g^{-1}\circ f^{-1}\right)(a)=\dfrac{a+20}{60} .
No.
Jika-
3-\dfrac1x -
4-\dfrac1x -
2-\dfrac1x
-
-2-\dfrac1x -
2+\dfrac1x
ALTERNATIF PENYELESAIAN
\begin{aligned}
(f\circ g)(x)&=\dfrac{x}{3x-2}\\[3.5pt]
f\left(g(x)\right)&=\dfrac{x}{3x-2}\\[3.5pt]
\dfrac1{2g(x)-1}&=\dfrac{x}{3x-2}\\[3.5pt]
2g(x)-1&=\dfrac{3x-2}{x}\\[3.5pt]
2g(x)-1&=3-\dfrac2x\\[3.5pt]
2g(x)&=4-\dfrac2x\\
g(x)&=\boxed{\boxed{2-\dfrac1x}}
\end{aligned}
Jadi, g(x)=2-\dfrac1x .
JAWAB: C
JAWAB: C
No.
Fungsi f dan g berikut adalah pemetaan dari R ke R. Tentukan rumus untuk fungsi komposisif(x) = 5x + 2 dang(x) = 4 − 2x f(x) = x2 + x dang(x) = x − 1 f(x) = x3 + x dang(x) = 2x2
ALTERNATIF PENYELESAIAN
(f∘g)(x) = f(g(x))
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(4-2x\right)\\ &=5\left(4-2x\right)+2\\ &=\boxed{\boxed{20-10x}} \end{aligned}\)
\(\begin{aligned} \left(g\circ f\right)(x)&=g\left(f(x)\right)\\ &=g(5x+2)\\ &=4-2(5x+2)\\ &=4-10x-4\\ &=\boxed{\boxed{-10x}} \end{aligned}\)(f∘g)(x) = f(g(x))
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(x-1\right)\\ &=\left(x-1\right)^2+x-1\\ &=x^2-2x+1+x-1\\ &=\boxed{\boxed{x^2-x}} \end{aligned}\)
\(\begin{aligned} \left(g\circ f\right)(x)&=g\left(f(x)\right)\\ &=g\left(x^2+x\right)\\ &=\boxed{\boxed{x^2+x-1}} \end{aligned}\)(f∘g)(x) = f(g(x))
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(2x^2\right)\\ &=\left(2x^2\right)^3+2x^2\\ &=\boxed{\boxed{8x^6+2x^2}} \end{aligned}\)
\(\begin{aligned} \left(g\circ f\right)(x)&=g\left(f(x)\right)\\ &=g\left(x^3+x\right)\\ &=2\left(x^3+x\right)^2\\ &=2\left(x^6+2x^4+x^2\right)\\ &=\boxed{\boxed{2x^6+4x^4+2x^2}} \end{aligned}\)
Jadi,
(f∘g)(x) = 20 − 10x dan(g∘f)(x) = −10x (f∘g)(x) = x2 − x dan(g∘f)(x) = x2 + x− 1 (f∘g)(x) = 8x6 + 2x2 dan(g∘f)(x) = 2x6 + 4x2 + 2x2
No.
Fungsi f dan g dinyatakan dalam bentuk pasangan terurut sebagai berikutNyatakan fungsi-fungsi komposisi berikut dalam pasangan terurut
- f ∘ g
- g ∘ f
- f ∘ g (5)
- f ∘ g (6)
- g ∘ f (−3)
- g ∘ f (0)
ALTERNATIF PENYELESAIAN
x −3 −2 −1 0 g 2 4 5 7 f ∘ g −2 −3 0 −1 f ∘ g : {(−3, −2), (−2, −3), (−1, 0), (0, −1)}
x 2 4 5 7 f −2 −3 0 −1 g ∘ f 4 2 7 5 f ∘ g : {(2, 4), (4, 2), (5, 7), (7, 5)} f ∘ g (5) = f (g(5))
Karena g(5) tidak ada, makaf ∘ g (5) tidak ada.
f ∘ g (6) = f (g(6))
Karena g(6) tidak ada, makaf ∘ g (6) tidak ada.
g ∘ f (−3) = g (f(−3))
Karena f(−3) tidak ada, makag ∘ f (−3) tidak ada.
g ∘ f (0) = g (f(0))
Karena f(0) tidak ada, makag ∘ f (0) tidak ada.
Jadi,
f ∘ g : {(−3, −2), (−2, −3), (−1, 0), (0, −1)} -
f ∘ g : {(2, 4), (4, 2), (5, 7), (7, 5)} f ∘ g (5) tidak adaf ∘ g (6) tidak adag ∘ f (−3) tidak adag ∘ f (0) tidak ada
No.
Fungsi f dan g berikut ini adalah pemetaan dari R ke R. Tentukanlah rumus untuk fungsi komposisif(x) = x − 1 dang(x) = 2x f(x) = 3x + 2 dang(x) = 4 − 3x f(x) = x2 dang(x) = 4x + 2 f(x) = 2x dang(x) = x2 − x f(x) = x2 − 1 dang(x) = x2 + 1
ALTERNATIF PENYELESAIAN
f(x) = x − 1 dang(x) = 2x
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(g(x)\right)\\ &=f\left(2x\right)\\ &=2x-1 \end{aligned}\)f(x) = 3x + 2 dang(x) = 4 − 3x
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(g(x)\right)\\ &=f\left(4-3x\right)\\ &=3(4-3x)+2\\ &=12-9x+2\\ &=14-9x \end{aligned}\)f(x) = x2 dang(x) = 4x + 2
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(g(x)\right)\\ &=f(4x+2)\\ &=(4x+2)^2\\ &=16x^2+16x+4 \end{aligned}\)f(x) = 2x dang(x) = x2 − x
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(g(x)\right)\\ &=f\left(x^2-x\right)\\ &=2\left(x^2-x\right)\\ &=2x^2-2x \end{aligned}\)f(x) = x2 − 1 dang(x) = x2 + 1
\(\begin{aligned} \left(f\circ g\right)(x)&=f\left(g(x)\right)\\ &=f\left(x^2+1\right)\\ &=\left(x^2+1\right)-1\\ &=x^4+2x^2+1-1\\ &=x^4+x^2 \end{aligned}\)
Jadi,
(f ∘ g)(x) = 2x − 1 (f ∘ g)(x) = 14 − 9x (f ∘ g)(x) = 16x2 + 16x + 4 (f ∘ g)(x) = 2x2 − 2x (f ∘ g)(x) = x4 + x2
No.
Fungsi- 6x + 3
- 6x − 3
- 6x + 5
- 6x − 5
- −6x + 5
ALTERNATIF PENYELESAIAN
\(\begin{aligned}
\left(f\circ g\right)(x)&=f\left(g(x)\right)\\
&=f\left(3x+2\right)\\
&=2(3x+2)+1\\
&=6x+4+1\\
&=\color{blue}\boxed{\boxed{\color{black}6x+5}}
\end{aligned}\)
Jadi, rumus fungsi (fˆg)(x) adalah 6x + 5.
JAWAB: C
JAWAB: C
No.
Diketahui- 2x2 − 2x − 3
- 2x2 + 2x − 1
- 4x2 − 2
- 4x2 − 4x − 2
- 4x2 − 4x − 4
ALTERNATIF PENYELESAIAN
\begin{aligned}
\left(f\circ g\right)(x)&=f\left(g(x)\right)\\
&=f\left(2x-1\right)\\
&=(2x-1)^2-3\\
&=4x^2-4x+1-3\\
&=\color{blue}\boxed{\boxed{\color{black}4x^2-4x-2}}
\end{aligned}
Jadi, (fˆg)(x) = 4x2 − 4x − 2.
JAWAB: D
JAWAB: D
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