Exercise Zone : Fungsi Komposisi

Berikut ini adalah kumpulan soal mengenai Fungsi Komposisi. Jika ingin bertanya soal, silahkan gabung ke grup Telegram, Signal, Discord, atau WhatsApp.

Tipe:

Standar SBMPTN HOTS

No.

Diketahui {f(x)=2x-3} dan {\left(g\circ f\right)(x)=2x+1}. Tentukan nilai g(x).

dari Nasywa

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

CARA BIASA	CARA CEPAT
Misal 2x-3=u $$\begin{array}{rl}2x& =u+3\\ x& ={\displaystyle \frac{u+3}{2}}\end{array}$$ $$\begin{array}{rl}g(2x-3)& =2x+1\\ g(u)& =2\left({\displaystyle \frac{u+3}{2}}\right)+1\\ & =u+3+1\\ & =u+4\\ g(x)& =\boxed{{\displaystyle \boxed{{\displaystyle x+4}}}}\end{array}$$	$$\begin{array}{rl}g(2x-3)& =2x+1\\ g(2x-3)& =2x-3+3+1\\ g(2x-3)& =2x-3+4\\ g(x)& =\boxed{{\displaystyle \boxed{{\displaystyle x+4}}}}\end{array}$$

Jadi, g(x)=x+4.

No.

Jika {f(x)=5x+3} dan {g(f(x))=4x+9}, nilai g(13) adalah....

1. 13
2. 14
   
3. 15
   

4. 16
5. 17
   

$$\begin{array}{rl}f(x)& =13\\ 5x+3& =13\\ 5x& =10\\ x& =2\end{array}$$

$$\begin{array}{rl}g(f(x))& =4x+9\\ g(13)& =4(2)+9\\ & =17\end{array}$$

Jadi, g(13)=17. JAWAB: E

No.

Diketahui {f(x)=2x+3} dan {\left(g\circ f\right)(x)=4x^2+16x+16}. Rumus fungsi g(x) adalah ....

1. x^2-2x
2. x^2-2x-1
   
3. x^2-2x+1
   

4. x^2+2x+1
5. x^2+2x-1
   

$$\begin{array}{rl}(g\circ f)(x)& =4x^2+16x+16\\ g(f(x))& =4x^2+16x+16\\ g(2x+3)& =4x^2+16x+16\end{array}$$

Misal t=2x+3
$$\begin{array}{rl}t-3& =2x\\ {\displaystyle \frac{t-3}{2}}& =x\\ x& ={\displaystyle \frac{t-3}{2}}\end{array}$$

$$\begin{array}{rl}g(t)& =4{\left({\displaystyle \frac{t-3}{2}}\right)}^2+16\left({\displaystyle \frac{t-3}{2}}\right)+16\\ & =4\left({\displaystyle \frac{t^2-6t+9}{4}}\right)+8(t-3)+16\\ & =t^2-6t+9+8t-24+16\\ & =t^2+2t+1\\ g(x)& =\boxed{{\displaystyle \boxed{{\displaystyle x^2+2x+1}}}}\end{array}$$

Jadi, g(x)=x^2+2x+1. JAWAB: D

No.

Jika {g(x)=x-2} dan {(g\circ f)(x)=x^2+2x+3}, maka (f\circ g)(3) adalah ....

1. 5
2. 6
   
3. 7
   

4. 8
5. 9
   

$$\begin{array}{rl}(f\circ g)(3)& =f(g(3))\\ & =f(3-2)\\ & =f(1)\end{array}$$

$$\begin{array}{rl}(g\circ f)(1)& =1^2+2(1)+3\\ g(f(1))& =1+2+3\\ f(1)-2& =6\\ f(1)& =8\\ (f\circ g)(3)& =\boxed{{\displaystyle \boxed{{\displaystyle 8}}}}\end{array}$$

Jadi, (f\circ g)(3)=8. JAWAB: D

No.

Diketahui {f:R\to R}, {g:R\to R}, {g(x)=2x+3} dan {\left(f\circ g\right)(x)=12x^2+32x+26}. Rumus f(x)= ....

1. {3x^2-2x+5}
2. {3x^2-2x+37}
   
3. {3x^2-2x+50}
   

4. {3x^2+2x-5}
5. {3x^2+2x-50}
   

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$$\begin{array}{rl}(f\circ g)(x)& =12x^2+32x+26\\ f(g(x))& =12x^2+32x+26\\ f(2x+3)& =12x^2+32x+26\end{array}$$

Misal
$$\begin{array}{rl}2x+3& =u\\ 2x& =u-3\\ x& ={\displaystyle \frac{u-3}{2}}\end{array}$$

$$\begin{array}{rl}f\left(u\right)& =12{\left({\displaystyle \frac{u-3}{2}}\right)}^2+32\left({\displaystyle \frac{u-3}{2}}\right)+26\\ & =12\left({\displaystyle \frac{u^2-6u+9}{4}}\right)+16(u-3)+26\\ & =3(u^2-6u+9)+16u-48+26\\ & =3u^2-18u+27+16u-22\\ & =3u^2-2u+5\\ f(x)& =\boxed{{\displaystyle \boxed{{\displaystyle 3x^2-2x+5}}}}\end{array}$$

Jadi, f(x)=3x^2-2x+5. JAWAB: A

No.

Jika g(x)=\dfrac{ax+2}{x+3} dan h(x)=\dfrac{5x-4}{-x+a}, nilai (g\circ h)(1)=2, maka nilai dari 3a adalah

1. 1
2. 2
3. 3

4. 4
5. 5

$$\begin{array}{rl}(g\circ h)(1)& =2\\ g(h(1))& =2\\ g\left({\displaystyle \frac{5(1)-4}{-1+a}}\right)& =2\\ g\left({\displaystyle \frac{1}{a-1}}\right)& =2\\ {\displaystyle \frac{a\left({\displaystyle \frac{1}{a-1}}\right)+2}{{\displaystyle \frac{1}{a-1}}+3}}& =2\\ {\displaystyle \frac{{\displaystyle \frac{a+2(a-1)}{a-1}}}{{\displaystyle \frac{1+3(a-1)}{a-1}}}}& =2\\ {\displaystyle \frac{{\displaystyle \frac{a+2a-2}{a-1}}}{{\displaystyle \frac{1+3a-3}{a-1}}}}& =2\\ {\displaystyle \frac{{\displaystyle \frac{3a-2}{a-1}}}{{\displaystyle \frac{3a-2}{a-1}}}}& =2\\ {\displaystyle \frac{3a-2}{3a-2}}& =2\\ 3a-2& =6a-4\\ 3a& =\boxed{{\displaystyle \boxed{{\displaystyle 2}}}}\end{array}$$

Jadi,3a=2. JAWAB: B

No.

Diketahui fungsi f(x)=5x+3 dan g(x)=x^2+ax+b. Jika \left(g\circ f\right)(1)=53 dan \left(g\circ f\right)(0)=8, maka nilai a+b adalah

1. 2
2. 3
   
3. 4
   

4. 5
5. 6
   

$$\begin{array}{rl}(g\circ f)(1)& =53\\ g(f(1))& =53\\ g(5(1)+3)& =53\\ g(8)& =53\\ 8^2+a(8)+b& =53\\ 64+8a+b& =53\\ 8a+b& =-11\end{array}$$

$$\begin{array}{rl}(g\circ f)(1)& =53\\ g(f(0))& =8\\ g(5(0)+3)& =8\\ g(3)& =8\\ 3^2+a(3)+b& =8\\ 9+3a+b& =8\\ 3a+b& =-1\end{array}$$

$$\begin{array}{rl}8a+b& =-11\\ 3a+b& =-1-\\ 5a& =-10\\ a& =-2\end{array}$$

$$\begin{array}{rl}3a+b& =-1\\ 3(-2)+b& =-1\\ -6+b& =-1\\ b& =5\end{array}$$

$$\begin{array}{rl}a+b& =-2+5\\ & =\boxed{{\displaystyle \boxed{{\displaystyle 3}}}}\end{array}$$

Jadi, a+b=3. JAWAB: B

No.

Jika f(x)=\dfrac3{2x-1} dan \left(f\circ g\right)(x)=\dfrac{3x+3}{x-1}, maka g(x-1)=

1. \dfrac{x+2}x, x\neq0
2. \dfrac{x-2}x, x\neq0
3. \dfrac{x+1}x, x\neq0

4. \dfrac{x-1}x, x\neq0
5. \dfrac{x}{x+1}, x\neq-1

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

Jadi, g(x-1)=\dfrac{x-1}x. JAWAB: D

No.

Diketahui {f(x)=x^2-5x+1} dan {g(x)=x-4}. Jika {\left(g\circ f\right)(a)=21}, maka a= ...

1. 3
2. 5
3. 6

4. 8
5. 9

SKB CPNS MATEMATIKA 2020

$$\begin{array}{rl}(g\circ f)(a)& =21\\ g(f(a))& =21\\ g(a^2-5a+1)& =21\\ a^2-5a+1-4& =21\\ a^2-5a-3& =21\\ a^2-5a-24& =0\\ (a-8)(a+3)& =0\end{array}$$

Jadi, a=8. JAWAB: D

No.

Diketahui {f (x) = 3x + p} dan {g (x) = 4x - 120} dengan (f\circ g) (x) = (g\circ f) (x),} maka f (20) =

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$$\begin{array}{rl}(f\circ g)(x)& =(g\circ f)(x)\\ f(g(x))& =g(f(x))\\ f(4x-120)& =g(3x+p)\\ 3(4x-120)+p& =4(3x+p)-120\\ 12x-360+p& =12x+4p-120\\ -360+p& =4p-120\\ -3p& =240\\ p& ={\displaystyle \frac{240}{-3}}\\ & =-80\end{array}$$

$$\begin{array}{rl}f(x)& =3x-80\\ f(20)& =3(20)-80\\ & =60-80\\ & =\boxed{{\displaystyle \boxed{{\displaystyle -20}}}}\end{array}$$

Jadi, f (20) =-20.