Exercise Zone : Fungsi Komposisi [2]

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Tipe:

Standar SBMPTN HOTS

No.

Diketahui f:R→ R dan g: R→ R dengan f(x) = \dfrac{x-4}{x+3} dan g(x) = 3x - 2, tentukan \left(f\circ g\right)^{-1} (x)!

brainly.co.id

\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+4
g(x) = x^2-4
h(x) = x-5
Tentukan \left(f\circ(g\circ h)\right)(x) ?​

Brainly.co.id

\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, \left(f\circ(g\circ h)\right)(x)=x^2-10x+25.

No.

Diketahui f(a)=12a+4 dan g(a)= 5a-2 tentukan \left(g^{-1}\circ f^{-1}\right)(a)​

Brainly.co.id

\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)(x)&=(f\circ g)^{-1}(x)\\ &=\boxed{\boxed{\dfrac{a+20}{60}}} \end{aligned}

Jadi, \left(g^{-1}\circ f^{-1}\right)(a)​=\dfrac{a+20}{60}.

No.

Jika f(x)=\dfrac1{2x-1} dan (f\circ g)(x)=\dfrac{x}{3x-2} maka nilai g(x) adalah ....

1. 3-\dfrac1x
2. 4-\dfrac1x
3. 2-\dfrac1x

4. -2-\dfrac1x
5. 2+\dfrac1x

\begin{aligned} (f\circ g)(x)&=\dfrac{x}{3x-2}\\[8pt] f\left(g(x)\right)&=\dfrac{x}{3x-2}\\[8pt] \dfrac1{2g(x)-1}&=\dfrac{x}{3x-2}\\[8pt] 2g(x)-1&=\dfrac{3x-2}{x}\\[8pt] 2g(x)-1&=3-\dfrac2x\\[8pt] 2g(x)&=4-\dfrac2x\\ g(x)&=\boxed{\boxed{2-\dfrac1x}} \end{aligned}

Jadi, g(x)=2-\dfrac1x. JAWAB: C