Pertanyaan

Diketahui f ( x ) = x + 1 2 x − 1 , g ( x ) = x 2 − 1 , dan h ( x ) = 3 x + 5 . Tentukan: d. ( h − 1 ∘ g − 1 ∘ f − 1 ) ( x ) ,

Diketahui $f (x) = \frac{2 x - 1}{x + 1}, g (x) = x^{2} - 1$ , dan $h (x) = 3 x + 5$ . Tentukan:

d. $(h^{- 1} \circ g^{- 1} \circ f^{- 1}) (x)$ ,

W. Lestari

Master Teacher

Mahasiswa/Alumni Universitas Sriwijaya

Jawaban terverifikasi

Jawaban

.

$open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals fraction numerator plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root minus 5 over denominator 3 end fraction$ .

Pembahasan

Diketahui , dan . Maka: - Menentukan - Menentukan - Menentukan Sehingga: Jadi, .

Diketahui $f open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction comma space g open parentheses x close parentheses equals x squared minus 1$ , dan h open parentheses x close parentheses equals 3 x plus 5 . Maka:

- Menentukan f to the power of negative 1 end exponent open parentheses x close parentheses

$space space space space space space f open parentheses x close parentheses equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction space space space space space space space space space space y equals fraction numerator 2 x minus 1 over denominator x plus 1 end fraction y open parentheses x plus 1 close parentheses equals 2 x minus 1 space space space x y plus y equals 2 x minus 1 space x y minus 2 x equals negative 1 minus y x open parentheses y minus 2 close parentheses equals negative 1 minus y space space space space space space space space space space x equals fraction numerator negative 1 minus y over denominator y minus 2 end fraction space space f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 1 minus x over denominator x minus 2 end fraction$

- Menentukan g to the power of negative 1 end exponent open parentheses x close parentheses

- Menentukan h to the power of negative 1 end exponent open parentheses x close parentheses

$space space space space space h open parentheses x close parentheses equals 3 x plus 5 space space space space space space space space space space y equals 3 x plus 5 space space space space y minus 5 equals 3 x space space space space space space space 3 x equals y minus 5 space space space space space space space space space x equals fraction numerator y minus 5 over denominator 3 end fraction h to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 5 over denominator 3 end fraction$

Sehingga:

$open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses f to the power of negative 1 end exponent open parentheses x close parentheses close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses fraction numerator negative 1 minus x over denominator x minus 2 end fraction close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent close parentheses open parentheses plus-or-minus square root of fraction numerator negative 1 minus x over denominator x minus 2 end fraction plus 1 end root close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent close parentheses open parentheses plus-or-minus square root of fraction numerator negative 1 minus x plus x minus 2 over denominator x minus 2 end fraction end root close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals open parentheses h to the power of negative 1 end exponent close parentheses open parentheses plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root close parentheses open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals fraction numerator plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root minus 5 over denominator 3 end fraction$

Jadi, $open parentheses h to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent close parentheses equals fraction numerator plus-or-minus square root of fraction numerator negative 3 over denominator x minus 2 end fraction end root minus 5 over denominator 3 end fraction$ .