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(x))^-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

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

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

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

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

Jawaban

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

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

cara 1

Jawaban

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

cara 2

Jawaban

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

cara 3

Jawaban

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

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

cara 1

Jawaban

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

cara 2

Jawaban

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

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

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

cara 2

Jawaban

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

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

cara 1

Jawaban

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

cara 2

Jawaban

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

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

cara 1

Jawaban

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

cara 2

Jawaban

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

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

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

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

Jawaban

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