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Diketahui , nilai dari  adalah ....

Pertanyaan

Diketahui p open parentheses x close parentheses equals fraction numerator 5 x minus 2 over denominator negative 4 x plus 2 end fraction, nilai dari p to the power of negative 1 end exponent open parentheses x close parentheses adalah ....

Pembahasan Soal:

Diketahui p open parentheses x close parentheses equals fraction numerator 5 x minus 2 over denominator negative 4 x plus 2 end fraction, maka:

space space space space space space space space space space space p open parentheses x close parentheses equals fraction numerator 5 x minus 2 over denominator negative 4 x plus 2 end fraction space space space space space space space space space space space space space space space space y equals fraction numerator 5 x minus 2 over denominator negative 4 x plus 2 end fraction y open parentheses negative 4 x plus 2 close parentheses equals 5 x minus 2 space minus 4 x y plus 2 y equals 5 x minus 2 space space minus 4 x y minus 5 x equals negative 2 minus 2 y space space x open parentheses negative 4 y minus 5 close parentheses equals negative 2 minus 2 y space space space space space space space space space space space space space space space space space x equals fraction numerator negative 2 minus 2 y over denominator negative 4 y minus 5 end fraction space space space space space space space space space p to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative 2 minus 2 x over denominator negative 4 x minus 5 end fraction 

Jadi, nilai dari p to the power of negative 1 end exponent open parentheses x close parentheses adalah fraction numerator negative 2 minus 2 x over denominator negative 4 x minus 5 end fraction.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

W. Lestari

Mahasiswa/Alumni Universitas Sriwijaya

Terakhir diupdate 06 Juni 2021

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Pertanyaan yang serupa

Jika , maka tentukan .

Pembahasan Soal:

Menentukan f to the power of negative 1 end exponent left parenthesis x right parenthesis

table attributes columnalign right center left columnspacing 0px end attributes row cell f left parenthesis x right parenthesis end cell equals cell fraction numerator 2 x minus 5 over denominator x minus 9 end fraction end cell row y equals cell fraction numerator 2 x minus 5 over denominator x minus 9 end fraction end cell row cell y left parenthesis x minus 9 right parenthesis end cell equals cell 2 x minus 5 end cell row cell x y minus 9 y end cell equals cell 2 x minus 5 end cell row cell x y minus 2 x end cell equals cell 9 y minus 5 end cell row cell x left parenthesis y minus 2 right parenthesis end cell equals cell 9 y minus 5 end cell row x equals cell fraction numerator 9 y minus 5 over denominator y minus 2 end fraction end cell row cell f to the power of negative 1 end exponent left parenthesis y right parenthesis end cell equals cell fraction numerator 9 y minus 5 over denominator y minus 2 end fraction end cell row cell f to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell fraction numerator 9 x minus 5 over denominator x minus 2 end fraction end cell end table 

f to the power of negative 1 end exponent left parenthesis x right parenthesis akan memiliki nilai jika penyebutnya tidak sama dengan nol, sehingga

table attributes columnalign right center left columnspacing 0px end attributes row cell x minus 2 end cell not equal to 0 row x not equal to 2 end table 

Jadi, invers dari f left parenthesis x right parenthesis adalah f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator 9 x minus 5 over denominator x minus 2 end fraction comma space x not equal to 2.

Roboguru

Jika  adalah invers dari , maka nilai

Pembahasan Soal:

Jika begin mathsize 14px style g to the power of negative 1 end exponent end style adalah invers dari begin mathsize 14px style g open parentheses x close parentheses equals fraction numerator 8 minus 3 x over denominator 4 minus x end fraction comma space x not equal to 4 end style, maka nilai begin mathsize 14px style g to the power of negative 1 end exponent open parentheses 4 close parentheses equals... end style 

begin mathsize 14px style g open parentheses x close parentheses equals fraction numerator 8 minus 3 x over denominator 4 minus x end fraction space space space space y equals fraction numerator 8 minus 3 x over denominator 4 minus x end fraction y open parentheses 4 minus x close parentheses equals 8 minus 3 x 4 y minus x y equals 8 minus 3 x 3 x minus x y equals 8 minus 4 y x open parentheses 3 minus y close parentheses equals 8 minus 4 y x equals fraction numerator 8 minus 4 y over denominator 3 minus y end fraction g to the power of negative 1 end exponent open parentheses y close parentheses equals fraction numerator 8 minus 4 y over denominator 3 minus y end fraction g to the power of negative 1 end exponent open parentheses 4 close parentheses equals fraction numerator 8 minus 4 open parentheses 4 close parentheses over denominator 3 minus 4 end fraction space space space space space space space space space space space space equals 8 end style 

 

Roboguru

Invers dari fungsi  adalah

Pembahasan Soal:

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 fraction numerator 2 x minus 7 over denominator 3 x plus 8 end fraction end cell row y equals cell fraction numerator 2 x minus 7 over denominator 3 x plus 8 end fraction end cell row cell 3 x y plus 8 y end cell equals cell 2 x minus 7 end cell row cell 3 x y minus 2 x end cell equals cell negative 8 y minus 7 end cell row cell x left parenthesis 3 y minus 2 right parenthesis end cell equals cell negative 8 y minus 7 end cell row x equals cell fraction numerator negative 8 y minus 7 over denominator 3 y minus 2 end fraction end cell row cell f to the power of negative 1 end exponent left parenthesis x right parenthesis end cell equals cell fraction numerator negative 8 x minus 7 over denominator 3 x minus 2 end fraction end cell end table end style

Roboguru

Jika , maka tentukan .

Pembahasan Soal:

misal space space fraction numerator 2 x minus 5 over denominator x minus 9 end fraction equals a space space space space space space space space space space space space 2 x minus 5 equals a open parentheses x minus 9 close parentheses space space space space space space space space space space space space 2 x minus 5 equals a x minus 9 a space space space space space space space space space 2 x minus a x equals negative 9 a plus 5 space space space space space space space space space open parentheses 2 minus a close parentheses x equals negative 9 a plus 5 space space space space space space space space space space space space space space space space space space space space x equals fraction numerator negative 9 a plus 5 over denominator 2 minus a end fraction space space space space space space space space space space space space space space space space space space space space x equals fraction numerator 9 a minus 5 over denominator a minus 2 end fraction 

Jadi, f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator 9 x minus 5 over denominator x minus 2 end fraction comma space x not equal to 2.

Roboguru

Jika , maka ....

Pembahasan Soal:

Ingat!

begin mathsize 14px style y equals f left parenthesis x right parenthesis left right arrow x equals f to the power of negative 1 end exponent left parenthesis y right parenthesis end style 

Maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row y equals cell f left parenthesis x right parenthesis end cell row y equals cell fraction numerator x minus 2 over denominator 2 end fraction end cell row cell 2 y end cell equals cell x minus 2 end cell row cell 2 y plus 2 end cell equals x end table end style 

Sehingga, begin mathsize 14px style f to the power of negative 1 end exponent left parenthesis y right parenthesis equals 2 y plus 2 end style atau begin mathsize 14px style f to the power of negative 1 end exponent left parenthesis x right parenthesis equals 2 x plus 2 end style 

Jadi, jawaban yang tepat adalah D.

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