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No.

Diketahui f(x) = 2x − 3 dan (g ∘ f)(x) = 2x + 1. Tentukan nilai g(x).
dari Nasywa
ALTERNATIF PENYELESAIAN
\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 BIASACARA 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\\[3.5pt] &=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
  1. 16
  2. 17
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN

CARA 1

\(\begin{aligned} g\left(f(x)\right)&=4x+9\\ g(5x+3)&=4x+9 \end{aligned}\)
Misal p = 5x + 3 ⟶ \(x=\dfrac{p-3}5\)
\(\begin{aligned} g(p)&=4\left(\dfrac{p-3}5\right)+9\\[4pt] &=\dfrac{4p-12}5+\dfrac{45}5\\[4pt] &=\dfrac{4p+33}5\\[4pt] g(13)&=\dfrac{4(13)+33}5\\[4pt] &=\dfrac{52+33}5\\[4pt] &=\dfrac{85}5\\ &=\boxed{\boxed{17}} \end{aligned}\)

CARA 2

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

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

No.

Diketahui f(x) = 2x + 3 dan (g ∘ f)(x) = 4x2 + 16x + 16. Rumus fungsi g(x) adalah ....
  1. x2 − 2x
  2. x2 − 2x − 1
  3. x2 − 2x + 1
  1. x2 + 2x + 1
  2. x2 + 2x − 1
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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}

  • CARA BIASA

    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}

  • CARA CEPAT

    \begin{aligned} g(2x+3)&=4x^2+12x+9+4x+7\\ &=4x^2+12x+9+4x+6+1\\ &=(2x+3)^2+2(2x+3)+1\\ g(x)&=\boxed{\boxed{x^2+2x+1}} \end{aligned}
Jadi, g(x) = x2 + 2x + 1.
JAWAB: D

No.

Jika g(x) = x − 2 dan (g ∘ f)(x) = x2 + 2x + 3, maka (f ∘ g)(3) adalah ....
  1. 5
  2. 6
  3. 7
  1. 8
  2. 9
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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 ∘ g)(3) = 8.
JAWAB: D

No.

Diketahui f : ℝ ⟶ ℝ, g : ℝ ⟶ ℝ, g(x) = 2x + 3 dan (f ∘ g)(x) = 12x2 + 32x + 26. Rumus f(x) = ....
  1. 3x2 − 2x + 5
  2. 3x2 − 2x + 37
  3. 3x2 − 2x + 50
  1. 3x2 + 2x − 5
  2. 3x2 + 2x − 50
Grup WhatsApp
ALTERNATIF PENYELESAIAN
\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}

CARA 1

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\\[3.5pt] &=12\left(\dfrac{u^2-6u+9}4\right)+16\left(u-3\right)+26\\[3.5pt] &=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}

CARA 2

\begin{aligned} f\left(2x+3\right)&=12x^2+36x+27-4x-1\\ &=3\left(4x^2+12x+9\right)-4x-6+5\\ &=3\left(2x+3\right)^2-2(2x+3)+5\\ f(x)&=\boxed{\boxed{3x^2-2x+5}} \end{aligned}
Jadi, f(x) = 3x2 − 2x + 5.
JAWAB: A

No.

Jika \(g(x)=\dfrac{ax+2}{x+3}\) dan \(h(x)=\dfrac{5x-4}{-x+a}\), nilai (g ∘ h)(1) = 2, maka nilai dari 3a adalah
  1. 1
  2. 2
  3. 3
  1. 4
  2. 5
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\begin{aligned} (g\circ h)(1)&=2\\ g(h(1))&=2\\ g\left(\dfrac{5(1)-4}{-1+a}\right)&=2\\[3.5pt] g\left(\dfrac1{a-1}\right)&=2\\[3.5pt] \dfrac{a\left(\dfrac1{a-1}\right)+2}{\dfrac1{a-1}+3}&=2\\[3.5pt] \dfrac{\dfrac{a+2(a-1)}{a-1}}{\dfrac{1+3(a-1)}{a-1}}&=2\\[3.5pt] \dfrac{\dfrac{a+2a-2}{a-1}}{\dfrac{1+3a-3}{a-1}}&=2\\[3.5pt] \dfrac{\dfrac{3a-2}{a-1}}{\dfrac{3a-2}{a-1}}&=2\\[3.5pt] \dfrac{3a-2}{3a-2}&=2\\[3.5pt] 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) = x2 + ax + b. Jika (g ∘ f)(1) = 53 dan (g ∘ f)(0) = 8, maka nilai a + b adalah
  1. 2
  2. 3
  3. 4
  1. 5
  2. 6
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\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&-\\\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 ≠ 0
  2. \(\dfrac{x-2}x\), x ≠ 0
  3. \(\dfrac{x+1}x\), x ≠ 0
  1. \(\dfrac{x-1}x\), x ≠ 0
  2. \(\dfrac{x}{x+1}\), x ≠ −1
Grup Matematika Idhamdaz
ALTERNATIF PENYELESAIAN
\begin{aligned} \left(f\circ g\right)(x)&=\dfrac{3x+3}{x-1}\\[3.5pt] f\left( g(x)\right)&=\dfrac{3x+3}{x-1}\\[3.5pt] \dfrac3{2g(x)-1}&=\dfrac{3x+3}{x-1}\\[3.5pt] \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}\\[3.5pt] &=\dfrac{x}{x+1}\\[3.5pt] 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) = x2 − 5x + 1 dan g(x) =x − 4. Jika (g ∘ f)(a) = 21, maka a = ...
  1. 3
  2. 5
  3. 6
  1. 8
  2. 9
SKB CPNS MATEMATIKA 2020
ALTERNATIF PENYELESAIAN
\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} a = 8 atau a = −3
Jadi, a = 8.
JAWAB: D

No.

Diketahui f(x) = 3x + p dan g(x) = 4x − 120 dengan (f ∘ g)(x) = (g ∘ f)(x), maka f(20) =
brainly.co.id
ALTERNATIF PENYELESAIAN
\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.


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