Exercise Zone : Fungsi Komposisi

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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{aligned} \left(g\circ f\right)(x)&=2x+1\\ g\left(f(x)\right)&=2x+1\\ g(2x-3)&=2x+1 \end{aligned}

CARA BIASA	CARA CEPAT
Misal 2x-3=u \begin{aligned} 2x&=u+3\\ x&=\dfrac{u+3}2 \end{aligned} \begin{aligned} g(2x-3)&=2x+1\\ g(u)&=2\left(\dfrac{u+3}2\right)+1\\[8pt] &=u+3+1\\ &=u+4\\ g(x)&=\boxed{\boxed{x+4}}\end{aligned}	\begin{aligned} g(2x-3)&=2x+1\\ g(2x-3)&=2x{\color{blue}{-3+3}}+1\\ g({\color{blue}{2x-3}})&={\color{blue}{2x-3}}+4\\ g(x)&=\boxed{\boxed{x+4}} \end{aligned}

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{aligned} f(x)&=13\\ 5x+3&=13\\ 5x&=10\\ x&=2 \end{aligned}

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

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{aligned} \left(g\circ f\right)(x)&=4x^2+16x+16\\ g(f(x))&=4x^2+16x+16\\ g(2x+3)&=4x^2+16x+16 \end{aligned}

Misal t=2x+3
\begin{aligned} t-3&=2x\\ \dfrac{t-3}2&=x\\ x&=\dfrac{t-3}2 \end{aligned}

\begin{aligned} g(t)&=4\left(\dfrac{t-3}2\right)^2+16\left(\dfrac{t-3}2\right)+16\\ &=4\left(\dfrac{t^2-6t+9}4\right)+8(t-3)+16\\ &=t^2-6t+9+8t-24+16\\ &=t^2+2t+1\\ g(x)&=\boxed{\boxed{x^2+2x+1}} \end{aligned}

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{aligned} (f\circ g)(3)&=f(g(3))\\ &=f(3-2)\\ &=f(1) \end{aligned}

\begin{aligned} (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{\boxed{8}} \end{aligned}

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}
   

Grup WhatsApp

\begin{aligned} \left(f\circ g\right)(x)&=12x^2+32x+26\\ f\left(g(x)\right)&=12x^2+32x+26\\ f\left(2x+3\right)&=12x^2+32x+26 \end{aligned}

Misal
\begin{aligned} 2x+3&=u\\ 2x&=u-3\\ x&=\dfrac{u-3}2 \end{aligned}

\begin{aligned} f\left(u\right)&=12\left(\dfrac{u-3}2\right)^2+32\left(\dfrac{u-3}2\right)+26\\[8pt] &=12\left(\dfrac{u^2-6u+9}4\right)+16\left(u-3\right)+26\\[8pt] &=3\left(u^2-6u+9\right)+16u-48+26\\ &=3u^2-18u+27+16u-22\\ &=3u^2-2u+5\\ f(x)&=\boxed{\boxed{3x^2-2x+5}} \end{aligned}

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{aligned} (g\circ h)(1)&=2\\ g(h(1))&=2\\ g\left(\dfrac{5(1)-4}{-1+a}\right)&=2\\[8pt] g\left(\dfrac1{a-1}\right)&=2\\[8pt] \dfrac{a\left(\dfrac1{a-1}\right)+2}{\dfrac1{a-1}+3}&=2\\[20pt] \dfrac{\dfrac{a+2(a-1)}{a-1}}{\dfrac{1+3(a-1)}{a-1}}&=2\\[20pt] \dfrac{\dfrac{a+2a-2}{a-1}}{\dfrac{1+3a-3}{a-1}}&=2\\[20pt] \dfrac{\dfrac{3a-2}{a-1}}{\dfrac{3a-2}{a-1}}&=2\\[20pt] \dfrac{3a-2}{3a-2}&=2\\[8pt] 3a-2&=6a-4\\ 3a&=\boxed{\boxed{2}} \end{aligned}

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{aligned} \left(g\circ f\right)(1)&=53\\ g\left(f(1)\right)&=53\\ g\left(5(1)+3\right)&=53\\ g(8)&=53\\ 8^2+a(8)+b&=53\\ 64+8a+b&=53\\ 8a+b&=-11 \end{aligned}

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

\begin{aligned} 8a+b&=-11\\ 3a+b&=-1\qquad-\\\hline 5a&=-10\\ a&=-2 \end{aligned}

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

\begin{aligned} a+b&=-2+5\\ &=\boxed{\boxed{3}} \end{aligned}

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{aligned} \left(f\circ g\right)(x)&=\dfrac{3x+3}{x-1}\\[8pt] f\left( g(x)\right)&=\dfrac{3x+3}{x-1}\\[8pt] \dfrac3{2g(x)-1}&=\dfrac{3x+3}{x-1}\\[8pt] \left(2g(x)-1\right)(3x+3)&=3(x-1)\\ 2(3x+3)g(x)-3x-3&=3x-3\\ (6x+6)g(x)&=6x\\ g(x)&=\dfrac{6x}{6x+6}\color{red}\dfrac{:6}{:6}\\[8pt] &=\dfrac{x}{x+1}\\[8pt] g(x-1)&=\dfrac{x-1}{x-1+1}\\ &=\boxed{\boxed{\dfrac{x-1}x}} \end{aligned}

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{aligned} \left(g\circ f\right)(a)&=21\\ g\left(f(a)\right)&=21\\ g\left(a^2-5a+1\right)&=21\\ a^2-5a+1-4&=21\\ a^2-5a-3&=21\\ a^2-5a-24&=0\\ (a-8)(a+3)&=0\end{aligned}

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) =

brainly.co.id

\begin{aligned} \left(f\circ g\right)(x)&=\left(g\circ f\right)(x)\\ f\left(g(x)\right)&=g\left(f(x)\right)\\ f\left(4x-120\right)&=g\left(3x+p\right)\\ 3\left(4x-120\right)+p&=4\left(3x+p\right)-120\\ 12x-360+p&=12x+4p-120\\ -360+p&=4p-120\\ -3p&=240\\ p&=\dfrac{240}{-3}\\ &=-80 \end{aligned}

\begin{aligned} f(x)&=3x-80\\ f(20)&=3(20)-80\\ &=60-80\\ &=\boxed{\boxed{-20}} \end{aligned}

Jadi, f (20) =-20.