f(x)=x2−1f(x)=x2+1 maka (f∘g)−1(x)+(g∘f)−1(x) adalah ....

Pertanyaan

$\style{font-size:14px}{f(x)=\sqrt{x^2-1}\\f(x)=\sqrt{x^2+1}}$

maka $\style{font-size:14px}{{\left(f\circ g\right)^{-1}(x)+}{(g\circ f)}^{-1}(x)}$ adalah ....

$begin mathsize 14px style 2 x end style$
$size 14px 2 size 14px x size 14px plus size 14px 1$
$begin mathsize 14px style x squared end style$
$begin mathsize 14px style x squared plus 1 end style$
$begin mathsize 14px style x to the power of 4 minus 1 end style$

Pembahasan Soal:

Pada soal yang diberikan, kedua fungsinya ditulis $begin mathsize 14px style f left parenthesis x right parenthesis end style$ seharusnya salah satunya adalah fungsi $size 14px g size 14px left parenthesis size 14px x size 14px right parenthesis$ . Kita misalkan fungsi yang kedua adalah $undefined$ , maka diperoleh:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell square root of x squared minus 1 end root end cell row y equals cell square root of x squared minus 1 end root end cell row cell y squared end cell equals cell x squared minus 1 end cell row cell x squared end cell equals cell y squared plus 1 end cell row x equals cell square root of y squared plus 1 end root end cell row cell f to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell square root of x squared plus 1 end root end cell row blank blank blank row cell g left parenthesis x right parenthesis end cell equals cell square root of x squared plus 1 end root end cell row y equals cell square root of x squared plus 1 end root end cell row cell y squared end cell equals cell x squared plus 1 end cell row cell x squared end cell equals cell y squared minus 1 end cell row x equals cell square root of y squared minus 1 end root end cell row cell g to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell square root of x squared minus 1 end root end cell end table end style$

sehingga diperoleh:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell open parentheses g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses left parenthesis x right parenthesis end cell row blank equals cell square root of open parentheses square root of x squared plus 1 end root close parentheses squared minus 1 end root end cell row blank equals cell square root of x squared plus 1 minus 1 end root end cell row blank equals x row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses left parenthesis x right parenthesis end cell row blank equals cell square root of open parentheses square root of x squared minus 1 end root close parentheses squared plus 1 end root end cell row blank equals cell square root of x squared minus 1 plus 1 end root end cell row blank equals x end table end style$

sehingga diperoleh:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis plus left parenthesis g ring operator f right parenthesis to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals x row blank equals cell 2 x end cell end table end style$

Jadi, jawaban yang benar adalah A.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

P. Afrisno

Mahasiswa/Alumni Universitas Sebelas Maret

Terakhir diupdate 05 Oktober 2021

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Pertanyaan yang serupa

Diketahui fungsi f(x)=3+2x, g(x)=x+43x−1,x=−4. Invers dari (f∘g)(x)= ...

Pembahasan Soal:

Tentukan terlebih dahulu komposisi fungsi $open parentheses f ring operator g close parentheses open parentheses x close parentheses$ , yaitu:

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell f open parentheses g open parentheses x close parentheses close parentheses end cell row blank equals cell f open parentheses fraction numerator 3 x minus 1 over denominator x plus 4 end fraction close parentheses end cell row blank equals cell 3 plus 2 open parentheses fraction numerator 3 x minus 1 over denominator x plus 4 end fraction close parentheses end cell row blank equals cell 3 plus fraction numerator 6 x minus 2 over denominator x plus 4 end fraction end cell row blank equals cell fraction numerator 3 open parentheses x plus 4 close parentheses over denominator x plus 4 end fraction plus fraction numerator 6 x minus 2 over denominator x plus 4 end fraction end cell row blank equals cell fraction numerator 3 x plus 12 over denominator x plus 4 end fraction plus fraction numerator 6 x minus 2 over denominator x plus 4 end fraction end cell row blank equals cell fraction numerator 9 x plus 10 over denominator x plus 4 end fraction end cell end table$

Kemudian tentukan invers dari $open parentheses f ring operator g close parentheses open parentheses x close parentheses$ .

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell fraction numerator 9 x plus 10 over denominator x plus 4 end fraction end cell row y equals cell fraction numerator 9 x plus 10 over denominator x plus 4 end fraction end cell row cell y open parentheses x plus 4 close parentheses end cell equals cell 9 x plus 10 end cell row cell x y plus 4 y end cell equals cell 9 x plus 10 end cell row cell x y minus 9 x end cell equals cell negative 4 y plus 10 end cell row cell x open parentheses y minus 9 close parentheses end cell equals cell negative 4 y plus 10 end cell row x equals cell fraction numerator negative 4 y plus 10 over denominator y minus 9 end fraction end cell row cell open parentheses f ring operator g close parentheses to the power of negative 1 end exponent end cell equals cell fraction numerator negative 4 x plus 10 over denominator x minus 9 end fraction end cell end table$

Maka, invers dari $open parentheses f ring operator g close parentheses open parentheses x close parentheses equals fraction numerator negative 4 x plus 10 over denominator x minus 9 end fraction$ .

Roboguru

Diketahui: f(x)=3x+4 dan g(x)=2x+14x−5;x=21. Tentukan invers dari f∘g(x)!

Pembahasan Soal:

Ingat bahwa:

Jika $undefined$ maka $begin mathsize 14px style f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator negative d x plus b over denominator c x minus a end fraction end style$

Komposisi $f (x) dan g (x)$ adalah

$f \circ g (x) = = = = = f (g (x)) 3 (\frac{4 x - 5}{2 x + 1}) + 4 \frac{12 x - 15}{2 x + 1} + 4 \frac{12 x - 15 + 8 x + 4}{2 x + 1} \frac{20 x - 11}{2 x + 1}$

Sehingga diperoleh inversnya adalah

$f \circ g^{- 1} (x) = = \frac{- x - 11}{2 x - 20} \frac{x + 11}{- 2 x + 20}$

Dengan demikian, invers dari $f \circ g (x)$ adalah $f \circ g^{- 1} (x) = \frac{x + 11}{- 2 x + 20}$ .

Roboguru

Jika f(x)=4x+6,g(x)=x−1x;x=1, dan h(x)=(f∘g)(x),h−1(x) adalah ...

Pembahasan Soal:

Dari soal diketahui

$f open parentheses x close parentheses equals 4 x plus 6 g open parentheses x close parentheses equals fraction numerator x over denominator x minus 1 end fraction comma x not equal to 1$

Fungsi $h open parentheses x close parentheses equals left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis$ dapat ditentukan dengan cara berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell h open parentheses x close parentheses end cell equals cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell row blank equals cell f open parentheses g open parentheses x close parentheses close parentheses end cell row blank equals cell f open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses end cell row blank equals cell 4 open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses plus 6 end cell row blank equals cell fraction numerator 4 x over denominator x minus 1 end fraction plus 6 end cell row blank equals cell fraction numerator 4 x plus 6 open parentheses x minus 1 close parentheses over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 4 x plus 6 x minus 6 over denominator x minus 1 end fraction end cell row blank equals cell fraction numerator 10 x minus 6 over denominator x minus 1 end fraction end cell end table$

Invers fungsi, merupakan fungsi kebalikan, sehingga $h to the power of negative 1 end exponent open parentheses x close parentheses$

$table attributes columnalign right center left columnspacing 0px end attributes row y equals cell fraction numerator 10 x minus 6 over denominator x minus 1 end fraction end cell row cell y x minus y end cell equals cell 10 x minus 6 end cell row cell y x minus 10 x end cell equals cell y minus 6 end cell row cell x open parentheses y minus 10 close parentheses end cell equals cell y minus 6 end cell row x equals cell fraction numerator y minus 6 over denominator y minus 10 end fraction end cell row cell h to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator x minus 6 over denominator x minus 10 end fraction comma x not equal to 10 end cell end table$

Oleh karena itu, jawaban yang tepat adalah A.

Roboguru

Diketahui fungsi f(x)=x−3x+2,x=3 dan g(x)=x+4. Invers fungsi (f∘g)(x) adalah ....

Pembahasan Soal:

Dengan menggunakan konsep fungsi komposisi, kemudian invers fungsi, maka didapatkan:

$(f \circ g) (x) y y x + y y x - x x (y - 1) x (f \circ g)^{- 1} (x) = = = = = = = = = f (g (x)) \frac{x + 4 + 2}{x + 4 - 3} \frac{x + 6}{x + 1} \frac{x + 6}{x + 1} x + 6 - y + 6 - y + 6 \frac{- y + 6}{y - 1} \frac{- x + 6}{x - 1}$

Jadi, invers fungsi $left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis$ adalah $table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator negative x plus 6 over denominator x minus 1 end fraction end cell end table$ .

Roboguru

Fungsi fdang ditentukan oleh f(x)=3x−1 dan g(x)=x−1x+2,x=1. Fungsi invers (g∘f) adalah (g∘f)−1. Nilai dari (g∘f)−1(−2)=....

Pembahasan Soal:

Diketahui:

$table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell 3 x minus 1 end cell row cell g open parentheses x close parentheses end cell equals cell fraction numerator x plus 2 over denominator x minus 1 end fraction comma space x not equal to 1 end cell end table$

Sehingga diperoleh $open parentheses g ring operator f close parentheses open parentheses x close parentheses$ sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g ring operator f close parentheses open parentheses x close parentheses end cell equals cell g open parentheses f open parentheses x close parentheses close parentheses end cell row blank equals cell fraction numerator f open parentheses x close parentheses plus 2 over denominator f open parentheses x close parentheses minus 1 end fraction end cell row blank equals cell fraction numerator 3 x minus 1 plus 2 over denominator 3 x minus 1 minus 1 end fraction end cell row blank equals cell fraction numerator 3 x plus 1 over denominator 3 x minus 2 end fraction end cell end table$

Selanjutnya dapat ditentukan invers fungsi $open parentheses g ring operator f close parentheses open parentheses x close parentheses$ sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g ring operator f close parentheses open parentheses x close parentheses end cell equals cell fraction numerator 3 x plus 1 over denominator 3 x minus 2 end fraction end cell row y equals cell fraction numerator 3 x plus 1 over denominator 3 x minus 2 end fraction end cell row cell y open parentheses 3 x minus 2 close parentheses end cell equals cell 3 x plus 1 end cell row cell 3 x y minus 2 y end cell equals cell 3 x plus 1 end cell row cell 3 x y minus 3 x end cell equals cell 2 y plus 1 end cell row cell x open parentheses 3 y minus 3 close parentheses end cell equals cell 2 y plus 1 end cell row x equals cell fraction numerator 2 y plus 1 over denominator 3 y minus 3 end fraction end cell row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator 2 x plus 1 over denominator 3 x minus 3 end fraction end cell end table$

Sehingga diperoleh nilai dari $Error converting from MathML to accessible text.$ sebagai berikut.

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell fraction numerator 2 x plus 1 over denominator 3 x minus 3 end fraction end cell row cell open parentheses g ring operator f close parentheses to the power of negative 1 end exponent open parentheses negative 2 close parentheses end cell equals cell fraction numerator 2 open parentheses negative 2 close parentheses plus 1 over denominator 3 open parentheses negative 2 close parentheses minus 3 end fraction end cell row blank equals cell fraction numerator negative 4 plus 1 over denominator negative 6 minus 3 end fraction end cell row blank equals cell fraction numerator negative 3 over denominator negative 9 end fraction end cell row blank equals cell 1 third end cell end table$

Jadi, jawaban yang benar adalah E.

Roboguru