Soal-Soal Matematika/Fungsi invers

Bentuk dan sifat fungsi invers

bentuk f^-1(x)

sifat:

1. y = f(x) → f^-1(y) = x
2. f(x-a) = g(x) → f(x) = g(x+a)
3. (f^-1)^-1(x) = f(x)
4. f(f^-1(x)) = f^-1(f(x)) = x
5. (f o f^-1) (x) = (f^-1 o f) (x) = x
6. (f o g)^-1 (x) = (g^-1 o f^-1) (x)
7. (f o g) (x) = h(x) → (f^-1 o h) (x) = g(x)
8. (f o g o h)^-1 (x) = (h^-1 o g^-1 o f^-1) (x)

Pembuktian y = f(x) diubah jadi f^-1(y) = x!

pembuktian

$$\begin{array}{cc}\text{misal }y=f(x)& =ax+b\\ y& =ax+b\\ x& =\frac{y-b}{a}\\ \frac{y-b}{a}\text{ adalah }{f}^{-1}(y)\\ x& =f^{-1}(y)\end{array}$$

Pembuktian bahwa f^-1(f(x)) = x!

sebelumnya pembuktian yang tadi (lihat diatas) lalu masukkan y = f(x) ke f^-1(y) = x menjadi f^-1(f(x)) = x.

tabel fungsi invers

Fungsi invers (f(x))	Hasil (f-1(x))
$ax+b$	$\frac{x-b}{a}$
$\frac{ax+b}{cx+d}$	$\frac{-dx+b}{cx-a}$
${(ax+b)}^2$	$\frac{\sqrt{x}-b}{a}$
$\sqrt{ax+b}$	$\frac{x^2-b}{a}$
$a^{(bx)}$	$\frac{{}^{a}logx}{b}$
${}^{a}logbx$	$\frac{a^x}{b}$

contoh soal

1. tentukan fungsi invers dari:

• f(x)=²log (x-7)
• f(3x-5)=x²-2x+9
• f($\frac{x+3}{x-6}$)=$\frac{4x+1}{x+3}$

Jawaban

$$\begin{array}{cc}\ast f(x)& {=}^{2}log(x-7)\\ y& {=}^{2}log(x-7)\\ 2^y& =x-7\\ x& =2^y+7\\ f^{-1}(x)& =2^x+7\\ \ast f(3x-5)& =x^2-2x+9\\ f(\frac{3x-5+5}{3})& =(\frac{x+5}{3}{)}^2-2(\frac{x+5}{3})+9\\ f(x)& =\frac{x^2+10x+25}{9}-(\frac{2x+10}{3})+9\\ & =\frac{x^2+10x+25-6x-30+81}{9}\\ & =\frac{x^2+4x+76}{9}\\ & =\frac{x^2+4x+4+72}{9}\\ & =\frac{(x+2{)}^2}{9}+8\\ y& =\frac{(x+2{)}^2}{9}+8\\ y-8& =\frac{(x+2{)}^2}{9}\\ 9(y-8)& =(x+2{)}^2\\ 9y-72& =(x+2{)}^2\\ \sqrt{9y-72}& =x+2\\ x& =\sqrt{9y-72}-2\\ & =\sqrt{9(y-8)}-2\\ & =3\sqrt{y-8}-2\\ f^{-1}(x)& =3\sqrt{x-8}-2\\ \ast f(\frac{x+3}{x-6})& =\frac{4x+1}{x+3}\\ f(\frac{x+3}{x+3-9})& =\frac{4(x+3)-11}{x+3}\\ \text{misalkan }x+3=a\text{ dan }x=a-3\\ f(\frac{a}{a-9})& =\frac{4a-11}{a}\\ & =4-\frac{11}{a}\\ \text{misalkan }\frac{a}{a-9}=b\text{ dan }a=\frac{9b}{b-1}\\ f(b)& =4-\frac{11}{\frac{9b}{b-1}}\\ & =4-\frac{11(b-1)}{9b}\\ & =\frac{36b-11b+11}{9b}\\ & =\frac{25b+11}{9b}\\ & =\frac{25}{9}+\frac{11}{9b}\\ y-\frac{25}{9}& =\frac{11}{9b}\\ 9b(y-\frac{25}{9})& =11\\ 9by-25b& =11\\ b(9y-25)& =11\\ b& =\frac{11}{9y-25}\\ f^{-1}(x)& =\frac{11}{9x-25}\end{array}$$

1. tentukan fungsi invers dari $f(x)=\frac{2x-1}{3x+4}$!

Jawaban

$$\begin{array}{cc}f(x)& =\frac{2x-1}{3x+4}\\ 3xf(x)+4f(x)& =2x-1\\ 4f(x)+1& =2x-3xf(x)\\ 4f(x)+1& =(2-3f(x))x\\ x& =\frac{4f(x)+1}{2-3f(x)}\\ f^{-1}(x)& =\frac{4x+1}{-3x+2}\\ f^{-1}(x)& =\frac{4x+1}{-(3x-2)}\\ f^{-1}(x)& =\frac{-4x-1}{3x-2}\end{array}$$

1. tentukan nilai f(1) jika $f^{-1}(x)=\frac{2x-1}{5x-7}$!

cara 1

Jawaban

$$\begin{array}{ccc}\text{misalkan }f(1)& =a\text{ maka }{f}^{-1}(a)& =1\\ \text{untuk }a& =x\\ f^{-1}(x)& =\frac{2x-1}{5x-7}\\ 1& =\frac{2x-1}{5x-7}\\ 5x-7& =2x-1\\ 3x& =6\\ x& =2\\ a& =x\\ a& =2\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}{f}^{-1}(x)& =\frac{2x-1}{5x-7}\\ \text{diubah menjadi }f(\frac{2x-1}{5x-7})& =x\\ f(\frac{2x-1}{5x-7})& =x\\ \text{anggap bahwa }f(1)& =f(\frac{2x-1}{5x-7})\\ 1& =\frac{2x-1}{5x-7}\\ 5x-7& =2x-1\\ 3x& =6\\ x& =2\\ f(\frac{2x-1}{5x-7})& =x\\ f(\frac{2(2)-1}{5(2)-7})& =2\\ f(\frac{3}{3})& =2\\ f(1)& =2\end{array}$$

cara 3

Jawaban

$$\begin{array}{cc}{f}^{-1}(x)& =\frac{2x-1}{5x-7}\\ \text{diubah menjadi }f(x)& =\frac{7x-1}{5x-2}\\ f(x)& =\frac{7x-1}{5x-2}\\ f(1)& =\frac{7(1)-1}{5(1)-2}\\ f(1)& =\frac{6}{3}\\ f(1)& =2\end{array}$$

1. tentukan nilai f^-1(1) jika $f(\frac{3}{2x-3})=\frac{2x+3}{x+4}$!

cara 1

Jawaban

$$\begin{array}{ccc}\text{misalkan}{f}^{-1}(1)& =a\text{ maka }f(a)& =1\\ \text{untuk }a& =\frac{3}{2x-3}\\ f(\frac{3}{2x-3})& =\frac{2x+3}{x+4}\\ 1& =\frac{2x+3}{x+4}\\ x+4& =2x+3\\ x& =1\\ a& =\frac{3}{2x-3}\\ a& =\frac{3}{2(1)-3}\\ a& =-3\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}f(\frac{3}{2x-3})& =\frac{2x+3}{x+4}\\ \text{diubah menjadi }{f}^{-1}(\frac{2x+3}{x+4})& =\frac{3}{2x-3}\\ f^{-1}(\frac{2x+3}{x+4})& =\frac{3}{2x-3}\\ \text{anggap bahwa }{f}^{-1}(1)& =f^{-1}(\frac{2x+3}{x+4})\\ 1& =\frac{2x+3}{x+4}\\ x+4& =2x+3\\ x& =1\\ f^{-1}(\frac{2x+3}{x+4})& =\frac{3}{2x-3}\\ f^{-1}(\frac{2(1)+3}{(1)+4})& =\frac{3}{2(1)-3}\\ f^{-1}(\frac{5}{5})& =\frac{3}{-1}\\ f^{-1}(1)& =-3\end{array}$$

1. Diketahui g(x)=x², (g ο f)(x)=x²-2x+1 dan h(x)=$\sqrt{5x-8}$. Tentukan (h^-1 ο g^-1 ο f^-1)(1)!

cara 1

Jawaban

$$\begin{array}{cc}g(x)& =x^2\\ g^{-1}(x)& =\sqrt{x}\\ (g\circ f)(x)& =x^2-2x+1\\ g(f(x))& =(x-1{)}^2\\ (f(x){)}^2& =(x-1{)}^2\\ f(x)& =x-1\\ f^{-1}(x)& =x+1\\ h(x)& =\sqrt{5x-8}\\ h^{-1}(x)& =\frac{x^2+8}{5}\\ (h^{-1}\circ g^{-1}\circ f^{-1})(1)& =h^{-1}(g^{-1}(f^{-1}(1)))\\ & =h^{-1}(g^{-1}((1)+1))\\ & =h^{-1}(g^{-1}(2))\\ & =h^{-1}(\sqrt{2})\\ & =\frac{(\sqrt{2}{)}^2+8}{5}\\ & =\frac{2+8}{5}\\ & =\frac{10}{5}\\ & =2\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}g(x)& =x^2\\ g^{-1}(x)& =\sqrt{x}\\ (g\circ f)(x)& =x^2-2x+1\\ g(f(x))& =(x-1{)}^2\\ (f(x){)}^2& =(x-1{)}^2\\ f(x)& =x-1\\ f^{-1}(x)& =x+1\\ h(x)& =\sqrt{5x-8}\\ h^{-1}(x)& =\frac{x^2+8}{5}\\ \text{karena }(h^{-1}\circ g^{-1}\circ f^{-1})(x)& =(f\circ g\circ h{)}^{-1}(x)\\ (f\circ g\circ h)(x)& =f(g(h(x)))\\ & =f(g(\sqrt{5x-8}))\\ & =f((\sqrt{5x-8}{)}^2)\\ & =f(5x-8)\\ & =5x-8-1\\ & =5x-9\\ \text{misal }(f\circ g\circ h)(x)& =y\\ y& =5x-9\\ x& =\frac{y+9}{5}\\ y^{-1}& =\frac{x+9}{5}\\ (f\circ g\circ h{)}^{-1}(x)& =\frac{x+9}{5}\\ (f\circ g\circ h{)}^{-1}(1)& =(f(g(h)){)}^{-1}(1)\\ & =\frac{1+9}{5}\\ & =\frac{10}{5}\\ & =2\end{array}$$

1. Diketahui $f^{-1}(\frac{1}{x-1})=\frac{x+3}{x-1}$. Tentukan nilai f(2)!

cara 1

Jawaban

$$\begin{array}{cc}{f}^{-1}(\frac{1}{x-1})& =\frac{x+3}{x-1}\\ \text{misalkan }f(2)& =p\\ f^{-1}(\frac{1}{x-1})& =\frac{x+3}{x-1}\text{ diubah menjadi }f(\frac{x+3}{x-1})=\frac{1}{x-1}\\ \text{perhatikan kedua persamaan dimana ruas kiri dan kanan sama hasilnya }\\ f(2)& =f(\frac{x+3}{x-1})\\ 2& =\frac{x+3}{x-1}\\ 2(x-1)& =x+3\\ 2x-2& =x+3\\ x& =5\\ p& =\frac{1}{x-1}\\ & =\frac{1}{5-1}\\ & =\frac{1}{4}\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}{f}^{-1}(\frac{1}{x-1})& =\frac{x+3}{x-1}\\ \text{misalkan }\frac{1}{x-1}& =p\text{ dan }x=\frac{1+p}{p}\\ f^{-1}(p)& =p(\frac{1+p}{p}+3)\\ & =p(\frac{1+p+3p}{p})\\ & =1+4p\\ \text{misalkan }{f}^{-1}(p)& =y\\ y& =1+4p\\ p& =\frac{y-1}{4}\\ f(p)& =\frac{p-1}{4}\\ f(2)& =\frac{2-1}{4}\\ & =\frac{1}{4}\end{array}$$

1. Diketahui $f^{-1}(\frac{3}{x+3})=\frac{2x+3}{x+3}$. Tentukan nilai a jika f(a)=1!

cara 1

Jawaban

$$\begin{array}{cc}{f}^{-1}(\frac{3}{x+3})& =\frac{2x+3}{x+3}\\ \text{serta }f(a)& =1\\ f^{-1}(\frac{3}{x+3})& =\frac{2x+3}{x+3}\text{ diubah menjadi }f(\frac{2x+3}{x+3})=\frac{3}{x+3}\\ \text{perhatikan kedua persamaan dimana ruas kiri dan kanan sama hasilnya }\\ 1& =\frac{3}{x+3}\\ x+3& =3\\ x& =0\\ a& =\frac{3}{x+3}\\ & =\frac{3}{0+3}\\ & =\frac{3}{3}\\ & =1\end{array}$$

cara 2

Jawaban

$$\begin{array}{cc}{f}^{-1}(\frac{3}{x+3})& =\frac{2x+3}{x+3}\\ f^{-1}(\frac{3}{x+3})& =\frac{3(\frac{2x}{3}+1)}{x+3}\\ \text{misalkan }\frac{3}{x+3}& =p\text{ dan }x=\frac{3-3p}{p}\\ f^{-1}(p)& =p(\frac{2(\frac{3-3p}{p})}{3}+1)\\ & =p(\frac{6-6p}{3p}+1)\\ & =p(\frac{6-6p+3p}{3p})\\ & =p(\frac{6-3p}{3p})\\ & =p(\frac{3(2-p)}{3p})\\ & =2-p\\ \text{misalkan }{f}^{-1}(p)& =y\\ y& =2-p\\ p& =2-y\\ f(p)& =2-p\\ f(a)& =2-a\\ f(a)& =1\\ 1& =2-a\\ a& =1\end{array}$$

1. Tentukan f^-1(x) jika:

f(x)=g(2x-3)

f(2x-1)=g(3x+1)

Jawaban

$$\begin{array}{cc}\ast f(x)=g(2x-3)& =y\\ f(x)& =y\\ f^{-1}(y)& =x\\ g(2x-3)& =y\\ g^{-1}(y)& =2x-3\\ & =2f^{-1}(y)-3\\ f^{-1}(y)& =\frac{g^{-1}(y)+3}{2}\\ f^{-1}(x)& =\frac{g^{-1}(x)+3}{2}\\ \ast f(2x-1)=g(3x+1)& =y\\ f(2x-1)& =y\\ f^{-1}(y)& =2x-1\\ x& =\frac{f^{-1}(y)+1}{2}\\ g(3x+1)& =y\\ g^{-1}(y)& =3x-1\\ g^{-1}(y)& =3(\frac{f^{-1}(y)+1}{2})-1\\ \frac{g^{-1}(y)+1}{3}& =\frac{f^{-1}(y)+1}{2}\\ f^{-1}(y)& =\frac{2(g^{-1}(y)+1)}{3}-1\\ & =\frac{2g^{-1}(y)-1}{3}\\ f^{-1}(x)& =\frac{2g^{-1}(x)-1}{3}\end{array}$$