Pertanyaan

Jika f ( x ) = 2 x + 2 , g ( x ) = 3 − x , dan h ( x ) = 4 x − 1 , maka ( g ∘ f ∘ h ) − 1 ( x ) yang tepat adalah....

Jika $f (x) = 2 x + 2, g (x) = 3 - x,$ dan $h (x) = 4 x - 1$ , maka $(g \circ f \circ h)^{- 1} (x)$ yang tepat adalah ....

$open parentheses g ring operator f ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 3 over denominator 8 end fraction$
$open parentheses g ring operator f ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 3 plus x over denominator 8 end fraction$
$open parentheses g ring operator f ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 3 minus x over denominator 8 end fraction$

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Jawaban terverifikasi

Jawaban

jawaban yang tepat adalah E.

Pembahasan

Cara 1 Tentukan terlebih dahulu. Kemudian, tentukan . Jika , maka didapat perhitungan sebagai berikut. Dengan demikian, . Cara 2 Kita cari masing-masing dan terlebih dahulu. Dimulai dari mencari . Perhatikan bahwa Kemudian, akan dicari . Perhatikan bahwa Selanjutnya, akan dicari . Perhatikan bahwa Ingat kembali bahwa , maka didapat hasil sebagai berikut. Jadi, jawaban yang tepat adalah E.

Cara 1

Tentukan terlebih dahulu.

Kemudian, tentukan .

Jika open parentheses g ring operator f ring operator h close parentheses open parentheses x close parentheses equals y , maka didapat perhitungan sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis g ring operator f ring operator h right parenthesis left parenthesis x right parenthesis end cell equals cell 3 minus 8 x end cell row y equals cell 3 minus 8 x end cell row cell 8 x end cell equals cell 3 minus y end cell row x equals cell fraction numerator 3 minus y over denominator 8 end fraction end cell row cell left parenthesis g ring operator f ring operator h right parenthesis to the power of negative 1 end exponent left parenthesis y right parenthesis end cell equals cell fraction numerator 3 minus y over denominator 8 end fraction end cell end table end style$

Dengan demikian, $begin mathsize 14px style open parentheses g ring operator f ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses equals table attributes columnalign right center left columnspacing 0px end attributes row blank blank cell fraction numerator 3 minus x over denominator 8 end fraction end cell end table end style$ .

Cara 2

Kita cari masing-masing dan terlebih dahulu.

Dimulai dari mencari f to the power of negative 1 end exponent . Perhatikan bahwa

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

Kemudian, akan dicari g to the power of negative 1 end exponent . Perhatikan bahwa

Selanjutnya, akan dicari h to the power of negative 1 end exponent . Perhatikan bahwa

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

Ingat kembali bahwa , maka didapat hasil sebagai berikut.

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses g ring operator f ring operator h close parentheses to the power of negative 1 end exponent open parentheses x close parentheses end cell equals cell open parentheses h to the power of negative 1 end exponent ring operator f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses end cell row blank equals cell h to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses g to the power of negative 1 end exponent open parentheses x close parentheses close parentheses close parentheses end cell row blank equals cell h to the power of negative 1 end exponent open parentheses f to the power of negative 1 end exponent open parentheses 3 minus x close parentheses close parentheses end cell row blank equals cell h to the power of negative 1 end exponent open parentheses fraction numerator open parentheses 3 minus x close parentheses minus 2 over denominator 2 end fraction close parentheses end cell row blank equals cell h to the power of negative 1 end exponent open parentheses fraction numerator 1 minus x over denominator 2 end fraction close parentheses end cell row blank equals cell fraction numerator open parentheses fraction numerator 1 minus x over denominator 2 end fraction close parentheses plus 1 over denominator 4 end fraction end cell row blank equals cell fraction numerator fraction numerator 1 minus x over denominator 2 end fraction plus 2 over 2 over denominator 4 end fraction end cell row blank equals cell fraction numerator fraction numerator 3 minus x over denominator 2 end fraction over denominator 4 end fraction end cell row blank equals cell fraction numerator 3 minus x over denominator 8 end fraction end cell end table end style$

Jadi, jawaban yang tepat adalah E.