Diketahui dan a. (f−1∘g−1)(x)?

Pertanyaan

Diketahui dan

a. $\style{font-size:14px}{(f^{-1}\circ g^{-1})(x)}$ ?

$\;$

P. Tessalonika

Master Teacher

Mahasiswa/Alumni Universitas Negeri Medan

Jawaban terverifikasi

Jawaban

diperoleh bahwa $begin mathsize 14px style open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals fraction numerator x plus 2 over denominator 2 end fraction end style$ .

Pembahasan

Dengan menggunakan rumus invers fungsi

$begin mathsize 14px style z open parentheses x close parentheses equals a x plus b blank semicolon a not equal to 0 z to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus b over denominator a end fraction blank semicolon a not equal to 0 end style$

Diperoleh

$begin mathsize 14px style f open parentheses x close parentheses equals x minus 3 f to the power of negative 1 end exponent open parentheses x close parentheses equals x plus 3 g open parentheses x close parentheses equals 2 x plus 4 g to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction end style$

Nilai adalah dimasukkan ke

$begin mathsize 14px style open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction plus 3 open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals fraction numerator x minus 4 over denominator 2 end fraction plus 6 over 2 open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals fraction numerator x plus 2 over denominator 2 end fraction end style$

Dengan demikian, diperoleh bahwa $begin mathsize 14px style open parentheses f to the power of negative 1 end exponent ring operator g to the power of negative 1 end exponent close parentheses open parentheses x close parentheses equals fraction numerator x plus 2 over denominator 2 end fraction end style$ .