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

$$\begin{array}{rl}(f\circ g)(x)& =f(g(x))\\ & =f(3x-2)\\ & ={\displaystyle \frac{3x-2-4}{3x-2+3}}\\ y& ={\displaystyle \frac{3x-6}{3x+1}}\\ (3x+1)y& =3x-6\\ 3xy+y& =3x-6\\ 3xy-3x& =-y-6\\ x(3y-3)& =-y-6\\ x& ={\displaystyle \frac{-y-6}{3y-3}}\\ (f\circ g{)}^{-1}(x)& ={\displaystyle \frac{-x-6}{3x-3}}\end{array}$$

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

$$\begin{array}{rl}(f\circ g\circ h)(x)& =f(g(h(x)))\\ & =f(g(x-5))\\ & =f((x-5{)}^2-4)\\ & =f(x^2-10x+25-4)\\ & =f(x^2-10x+21)\\ & =x^2-10x+21+4\\ & =\boxed{{\displaystyle \boxed{{\displaystyle x^2-10x+25}}}}\end{array}$$

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

$$\begin{array}{rl}(f\circ g)(x)& =f(g(x))\\ & =f(5a-2)\\ & =12(5a-2)+4\\ & =60a-24+4\\ & =60a-20\end{array}$$

$$\begin{array}{rl}(g^{-1}\circ f^{-1})(x)& =(f\circ g{)}^{-1}(x)\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{a+20}{60}}}}}}\end{array}$$

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

$$\begin{array}{rl}(f\circ g)(x)& ={\displaystyle \frac{x}{3x-2}}\\ f(g(x))& ={\displaystyle \frac{x}{3x-2}}\\ {\displaystyle \frac{1}{2g(x)-1}}& ={\displaystyle \frac{x}{3x-2}}\\ 2g(x)-1& ={\displaystyle \frac{3x-2}{x}}\\ 2g(x)-1& =3-{\displaystyle \frac{2}{x}}\\ 2g(x)& =4-{\displaystyle \frac{2}{x}}\\ g(x)& =\boxed{{\displaystyle \boxed{{\displaystyle 2-{\displaystyle \frac{1}{x}}}}}}\end{array}$$

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