Soal-Soal Matematika/Fungsi komposisi

Bentuk dan sifat fungsi komposisi

bentuk (f o g) x = f(g (x))

sifat:

1. (f o g) x ≠ (g o f) x
2. ((f o g) o h) x = (f o (g o h)) x
3. (f o I) x = (I o f) x

tambahan sifat-sifat dari fungsi invers.

contoh soal

1. f(x) = 6x-7, g(x) = 8x+1 serta h(x) = 5x-3. tentukan

• (f o g) x
• (g o f) x
• ((f o g) o h) x
• (f o (g o h)) x

Jawaban

$$\begin{array}{cc}\ast (f\circ g)x& =f(g(x))\\ & =6(8x+1)-7\\ & =48x+6-7\\ & =48x-1\\ \ast (g\circ f)x& =g(f(x))\\ & =8(6x-7)+1\\ & =48x-56+1\\ & =48x-55\\ \ast (f\circ g)x& =f(g(x))\\ & =6(8x+1)-7\\ & =48x+6-7\\ & =48x-1\\ ((f\circ g)\circ h)x& =(f\circ g)(h(x))\\ & =48(5x-3)-1\\ & =240x-144-1\\ & =240x-145\\ \ast (g\circ h)x& =g(h(x))\\ & =8(5x-3)+1\\ & =40x-24+1\\ & =40x-23\\ (f\circ (g\circ h))x& =f((g\circ h)(x))\\ & =6(40x-23)-7\\ & =240x-138-7\\ & =240x-145\end{array}$$

2. f(x) = 6x-7, g(x) = 4x²-8x+1. tentukan

• (f o g) x
• (g o f) x

Jawaban

$$\begin{array}{cc}\ast (f\circ g)x& =f(g(x))\\ & =6(4x^2-8x+1)-7\\ & =24x^2-48x+6-7\\ & =24x^2-48x-1\\ \ast (g\circ f)x& =g(f(x))\\ & =4(6x-7{)}^2-8(6x-7)+1\\ & =4(36x^2-84x+49)-48x+56+1\\ & =144x^2-336x+196-48x+57\\ & =144x^2-384x+253\end{array}$$

3. tentukan g(x) bila f(x) = 6x-7 serta

• f(g(x)) = 24x²-48x-1!
• g(f(x)) = 144x²-384x+253!

Jawaban

$$\begin{array}{cc}\ast f(g(x))& =24x^2-48x-1\\ 6(g(x))-7& =24x^2-48x-1\\ \frac{6(g(x))-7+7}{6}& =\frac{24x^2-48x-1+7}{6}\\ g(x)& =4x^2-8x+1\\ \ast g(f(x))& =144x^2-384x+253\\ g(6x-7)& =144x^2-384x+253\\ g(\frac{6x-7+7}{6})& =144(\frac{x+7}{6}{)}^2-384(\frac{x+7}{6})+253\\ g(x)& =\frac{144(x^2+14x+49)}{36}-\frac{(384x+2.688)}{6}+253\\ & =\frac{24(x^2+14x+49)}{6}-\frac{(384x+2.688)}{6}+\frac{1.518}{6}\\ g(x)& =\frac{24x^2+336x+1.176-384x-2.688+1.518}{6}\\ & =\frac{24x^2-48x+6}{6}\\ & =4x^2-8x+1\end{array}$$

4. tentukan f(x) bila g(x) = 4x²-8x+1 serta

• f(g(x)) = 24x²-48x-1!
• g(f(x)) = 144x²-384x+253!

Jawaban

$$\begin{array}{cc}\ast f(g(x))& =24x^2-48x-1\\ f(4x^2-8x+1)& =24x^2-48x-1\\ & =24x^2-48x-1+7-7\\ & =24x^2-48x+6-7\\ & =6(4x^2-8x+1)-7\\ f(x)& =6x-7\\ \ast g(f(x))& =144x^2-384x+253\\ 4f(x{)}^2-8f(x)+1& =144x^2-384x+253\\ 4f(x{)}^2-8f(x)+4-3& =144x^2-384x+256-3\\ (2f(x)-2{)}^2-3& =(12x-16{)}^2-3\\ (2f(x)-2{)}^2& =(12x-16{)}^2\\ 2f(x)-2& =12x-16\\ 2(f(x)-1)& =2(6x-8)\\ f(x)-1& =6x-8\\ f(x)& =6x-7\end{array}$$

5 Diketahui f(x)=2x+1 dan (f ο g)(x-1) = -2x²+4x-1. Tentukan g(-2)!

Jawaban

$$\begin{array}{cc}(f\circ g)(x-1)& =-2x^2+4x-1\\ f(g(x-1))& =-2x^2+4x-1\\ \text{misalkan }g(x-1)& =a\\ f(a)& =-2x^2+4x-1\\ 2a+1& =-2x^2+4x-1\\ 2a& =-2x^2+4x-2\\ a& =-x^2+2x-1\\ & =-(x^2-2x+1)\\ & =-(x-1{)}^2\\ g(x-1)& =a\\ & =-(x-1{)}^2\\ g(x)& =-x^2\\ g(-2)& =-(-2{)}^2\\ & =-4\end{array}$$

6 Diketahui f(x-4)=2x+1 dan g(x+3) = x²-5x+6. Tentukan (f ο g)^-1(9)!

cara 1

Jawaban

$$\begin{array}{cc}(f\circ g{)}^{-1}(9)& =g^{-1}(f^{-1}(9))\\ f(x-4)& =2x+1\\ f(x)& =2(x+4)+1\\ & =2x+8+1\\ & =2x+9\\ f^{-1}(x)& =\frac{x-9}{2}\\ g(x+3)& =x^2-5x+6\\ g(x)& =(x-3{)}^2-5(x-3)+6\\ & =x^2-6x+9-5x+15+6\\ & =x^2-11x+30\\ & =x^2-11x+\frac{121}{4}-\frac{1}{4}\\ & =(x-\frac{11}{2}{)}^2-\frac{1}{4}\\ g^{-1}(x)& =\sqrt{x+\frac{1}{4}}+\frac{11}{2}\\ (f\circ g{)}^{-1}(x)& =g^{-1}(f^{-1}(x))\\ g^{-1}(f^{-1}(x))& =\sqrt{\frac{x-9}{2}+\frac{1}{4}}+\frac{11}{2}\\ g^{-1}(f^{-1}(9))& =\sqrt{\frac{9-9}{2}+\frac{1}{4}}+\frac{11}{2}\\ & =\sqrt{\frac{1}{4}}+\frac{11}{2}\\ & =\frac{1}{2}+\frac{11}{2}\\ & =\frac{12}{2}\\ & =6\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}(f\circ g{)}^{-1}(9)& =g^{-1}(f^{-1}(9))\\ f(x-4)& =2x+1\\ f(x)& =2(x+4)+1\\ & =2x+8+1\\ & =2x+9\\ g(x+3)& =x^2-5x+6\\ g(x)& =(x-3{)}^2-5(x-3)+6\\ & =x^2-6x+9-5x+15+6\\ & =x^2-11x+30\\ & =x^2-11x+\frac{121}{4}-\frac{1}{4}\\ & =(x-\frac{11}{2}{)}^2-\frac{1}{4}\\ (f\circ g)(x)& =f(g(x))\\ & =2((x-\frac{11}{2}{)}^2-\frac{1}{4})+9\\ & =2(x-\frac{11}{2}{)}^2-\frac{1}{2}+9\\ & =2(x-\frac{11}{2}{)}^2+\frac{17}{2}\\ (f\circ g{)}^{-1}(x)& =\sqrt{\frac{x}{2}-\frac{17}{4}}+\frac{11}{2}\\ (f\circ g{)}^{-1}(9)& =\sqrt{\frac{9}{2}-\frac{17}{4}}+\frac{11}{2}\\ & =\frac{1}{2}+\frac{11}{9}\\ & =\frac{12}{2}\\ & =2\end{array}$$

7 Berapakah hasilnya f(g(-2)) + g(f(4)) jika f(5x-1) = x dan g(7-x) = x?

cara 1

Jawaban

$$\begin{array}{cc}f(5x-1)& =x\\ f(4)& =f(5(1)-1)\\ & =1\\ f(9)& =f(5(2)-1)\\ & =2\\ g(7-x)& =x\\ g(-2)& =g(7-9)\\ & =9\\ g(1)& =g(7-6)\\ & =6\\ f(g(-2))+g(f(4))& =f(9)+g(1)\\ & =2+6\\ & =8\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}f(5x-1)& =x\\ f(x)& =\frac{x+1}{5}\\ f(4)& =\frac{4+1}{5}\\ & =1\\ f(9)& =\frac{9+1}{5}\\ & =2\\ g(7-x)& =x\\ g(x)& =-x+7\\ g(-2)& =-(-2)+7\\ & =9\\ g(1)& =-1+7\\ & =6\\ f(g(-2))+g(f(4))& =f(9)+g(1)\\ & =2+6\\ & =8\end{array}$$

8 Diketahui g(x) = 1-x² dan f(g(x)) = $\frac{2-x^2}{x^2}$. Tentukan:

• $f(\frac{1}{3})+f(\frac{1}{9})$
• $f(\frac{1}{3})-f(\frac{1}{9})$
• $f(\frac{1}{3})\cdot f(\frac{1}{9})$
• $\frac{f(\frac{1}{3})}{f(\frac{1}{9})}$

cara 1

Jawaban

$$\begin{array}{cc}g(x)& =\frac{1}{3}\\ 1-x^2& =\frac{1}{3}\\ x^2& =1-\frac{1}{3}\\ & =\frac{2}{3}\\ f(\frac{1}{3})& =\frac{2-\frac{2}{3}}{\frac{2}{3}}\\ & =\frac{\frac{4}{3}}{\frac{2}{3}}\\ & =2\\ g(x)& =\frac{1}{9}\\ 1-x^2& =\frac{1}{9}\\ x^2& =1-\frac{1}{9}\\ & =\frac{8}{9}\\ f(\frac{1}{9})& =\frac{2-\frac{8}{9}}{\frac{8}{9}}\\ & =\frac{\frac{10}{9}}{\frac{8}{9}}\\ & =\frac{5}{4}\\ \ast f(\frac{1}{3})+f(\frac{1}{9})& =2+\frac{5}{4}\\ & =\frac{13}{4}\\ \ast f(\frac{1}{3})-f(\frac{1}{9})& =2-\frac{5}{4}\\ & =\frac{3}{4}\\ \ast f(\frac{1}{3})\cdot f(\frac{1}{9})& =2\cdot \frac{5}{4}\\ & =\frac{5}{2}\\ \ast \frac{f(\frac{1}{3})}{f(\frac{1}{9})}& =\frac{2}{\frac{5}{4}}\\ & =\frac{8}{5}\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}f(g(x))& =\frac{2-x^2}{x^2}\\ f(1-x^2)& =\frac{2-x^2}{x^2}\\ \text{misalkan }1-x^2& =y\text{ dan }{x}^2=1-y\\ f(1-x^2)& =\frac{2-x^2}{x^2}\\ f(y)& =\frac{2-(1-y)}{1-y}\\ & =\frac{1+y}{1-y}\\ f(\frac{1}{3})& =\frac{1+y}{1-y}\\ & =\frac{1+\frac{1}{3}}{1-\frac{1}{3}}\\ & =\frac{\frac{4}{3}}{\frac{2}{3}}\\ & =2\\ f(\frac{1}{9})& =\frac{1+y}{1-y}\\ & =\frac{1+\frac{1}{9}}{1-\frac{1}{9}}\\ & =\frac{\frac{10}{9}}{\frac{8}{9}}\\ & =\frac{5}{4}\\ \ast f(\frac{1}{3})+f(\frac{1}{9})& =2+\frac{5}{4}\\ & =\frac{13}{4}\\ \ast f(\frac{1}{3})-f(\frac{1}{9})& =2-\frac{5}{4}\\ & =\frac{3}{4}\\ \ast f(\frac{1}{3})\cdot f(\frac{1}{9})& =2\cdot \frac{5}{4}\\ & =\frac{5}{2}\\ \ast \frac{f(\frac{1}{3})}{f(\frac{1}{9})}& =\frac{2}{\frac{5}{4}}\\ & =\frac{8}{5}\end{array}$$