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Diketahui f(x)=x+3 dan (g∘f)(x)=2x2+4x−3. Nilai dari g(2) adalah ....

Pertanyaan

Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals x plus 3 end style dan begin mathsize 14px style left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis equals 2 x squared plus 4 x minus 3 end style. Nilai dari begin mathsize 14px style g left parenthesis 2 right parenthesis end style adalah ....

Pembahasan Soal:

Ingat bahwa, begin mathsize 14px style left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis equals g left parenthesis f left parenthesis x right parenthesis right parenthesis end style dan begin mathsize 14px style f left parenthesis f to the power of negative 1 end exponent left parenthesis x right parenthesis right parenthesis equals x end style, maka 

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

Carilah invers fungsi dari begin mathsize 14px style f end style. Diketahui begin mathsize 14px style f left parenthesis x right parenthesis equals x plus 3 end style, maka

begin mathsize 14px style rightwards arrow y equals x plus 3 rightwards arrow x equals y minus 3 rightwards arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals x minus 3 end style

 begin mathsize 14px style left parenthesis g ring operator f right parenthesis left parenthesis x right parenthesis equals 2 x squared plus 4 x minus 3 end style. maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g left parenthesis x right parenthesis end cell equals cell g left parenthesis f left parenthesis f to the power of negative 1 end exponent left parenthesis x right parenthesis right parenthesis right parenthesis end cell row blank equals cell open parentheses g ring operator f close parentheses left parenthesis f to the power of negative 1 end exponent left parenthesis x right parenthesis right parenthesis end cell row blank equals cell 2 left parenthesis x minus 3 right parenthesis squared plus 4 left parenthesis x minus 3 right parenthesis minus 3 end cell row blank equals cell 2 left parenthesis x squared minus 6 x plus 9 right parenthesis plus 4 x minus 12 minus 3 end cell row blank equals cell 2 x squared minus 12 x plus 18 plus 4 x minus 15 end cell row blank equals cell 2 x squared minus 8 x plus 3 end cell end table end style

maka

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell g left parenthesis 2 right parenthesis end cell equals cell 2 left parenthesis 2 right parenthesis squared minus 8 left parenthesis 2 right parenthesis plus 3 end cell row blank equals cell 8 minus 16 plus 3 end cell row blank equals cell negative 5 end cell end table end style

Jadi nilai dari begin mathsize 14px style g left parenthesis 2 right parenthesis equals negative 5 end style.

Pembahasan terverifikasi oleh Roboguru

Dijawab oleh:

F. Kurnia

Mahasiswa/Alumni Universitas Jember

Terakhir diupdate 05 Oktober 2021

Roboguru sudah bisa jawab 91.4% pertanyaan dengan benar

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Diketahui fungsi g(x)=2x−4 dan (f∘g)(x)=5x−97x+3​. Nilai dari f(2)....

Pembahasan Soal:

Dengan menggunakan konsep fungsi komposisi, didapatkan

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell equals cell fraction numerator 7 x plus 3 over denominator 5 x minus 9 end fraction end cell row cell f open parentheses 2 x minus 4 close parentheses end cell equals cell fraction numerator 7 x plus 3 over denominator 5 x minus 9 end fraction end cell row blank blank blank row cell 2 x minus 4 end cell equals a row cell 2 x end cell equals cell a plus 4 end cell row x equals cell fraction numerator a plus 4 over denominator 2 end fraction end cell row blank blank blank row cell f left parenthesis a right parenthesis end cell equals cell fraction numerator 7 open parentheses fraction numerator a plus 4 over denominator 2 end fraction close parentheses plus 3 over denominator 5 open parentheses fraction numerator a plus 4 over denominator 2 end fraction close parentheses minus 9 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 7 a plus 28 over denominator 2 end fraction end style plus 3 over denominator begin display style fraction numerator 5 a plus 20 over denominator 2 end fraction end style minus 9 end fraction end cell row blank equals cell fraction numerator begin display style fraction numerator 7 a plus 34 over denominator 2 end fraction end style over denominator begin display style fraction numerator 5 a plus 2 over denominator 2 end fraction end style end fraction end cell row blank equals cell fraction numerator 7 a plus 34 over denominator 5 a plus 2 end fraction end cell row blank blank blank row cell f left parenthesis 2 right parenthesis end cell equals cell fraction numerator 7 left parenthesis 2 right parenthesis plus 34 over denominator 5 left parenthesis 2 right parenthesis plus 2 end fraction end cell row blank equals cell fraction numerator 14 plus 34 over denominator 10 plus 2 end fraction end cell row blank equals cell 48 over 12 end cell row blank equals 4 end table end style

Jadi, nilai dari begin mathsize 14px style f left parenthesis 2 right parenthesis equals 4 end style.

0

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Diketahui f(2x−3)=5x+1. Tentukan: b. f(5)+f(3)

Pembahasan Soal:

Diketahui:

f left parenthesis 2 x minus 3 right parenthesis equals 5 x plus 1

Ditanyakan:

f open parentheses 5 close parentheses plus f open parentheses 3 close parentheses equals... ?

Penyelesaian:

Pertama kita tentukan rumus f(x) terlebih dahulu:

misalkan:

table row cell u equals 2 x minus 3 rightwards arrow end cell row blank row blank row blank end table table row cell 2 x minus 3 equals u end cell row cell 2 x equals u plus 3 end cell row cell x equals fraction numerator u plus 3 over denominator 2 end fraction end cell end table  

Sehingga:

f left parenthesis 2 x minus 3 right parenthesis equals 5 x plus 1 f open parentheses u close parentheses equals 5 open parentheses fraction numerator u plus 3 over denominator 2 end fraction close parentheses plus 1 f open parentheses u close parentheses equals fraction numerator 5 u plus 15 over denominator 2 end fraction plus 1 f open parentheses u close parentheses equals fraction numerator 5 u plus 15 plus 2 over denominator 2 end fraction f open parentheses x close parentheses equals fraction numerator 5 x plus 17 over denominator 2 end fraction 

Sehingga,

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses 5 close parentheses plus f open parentheses 3 close parentheses end cell equals cell fraction numerator 5 open parentheses 5 close parentheses plus 17 over denominator 2 end fraction plus fraction numerator 5 open parentheses 3 close parentheses plus 17 over denominator 2 end fraction end cell row blank equals cell fraction numerator 25 plus 17 over denominator 2 end fraction plus fraction numerator 15 plus 17 over denominator 2 end fraction end cell row blank equals cell 42 over 2 plus 32 over 2 end cell row blank equals 37 end table

Jadi, hasil dari f open parentheses 5 close parentheses plus f open parentheses 3 close parentheses adalah 37

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Diketahui fungsi f:R→R dan g:R→R. Jika diketahui (f∘g)(x)=x3−6x2+10x−3 dan g(x)=x−2, nilai dari f(2) adalah ....

Pembahasan Soal:

begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell left parenthesis f ring operator g right parenthesis left parenthesis x right parenthesis end cell equals cell x cubed minus 6 x squared plus 10 x minus 3 end cell row cell f left parenthesis x minus 2 right parenthesis end cell equals cell x cubed minus 6 x squared plus 10 x minus 3 end cell row blank blank blank row cell x minus 2 end cell equals 2 row x equals 4 row blank blank blank row cell f left parenthesis 2 right parenthesis end cell equals cell 4 cubed minus 6 left parenthesis 4 right parenthesis squared plus 10 left parenthesis 4 right parenthesis minus 3 end cell row blank equals cell 64 minus 96 plus 40 minus 3 end cell row blank equals 5 end table end style

Jadi, jawaban yang tepat adalah D.

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Diketahui fungsi f:R→R dan g:R→R dengan g(x)=−x+3 dan (f∘g)(x)=4x2−26x+32, maka nilai f(1) adalah ....

Pembahasan Soal:

Diketahui:

table attributes columnalign right center left columnspacing 0px end attributes row cell g left parenthesis x right parenthesis end cell equals cell negative x plus 3 end cell row cell open parentheses f ring operator g close parentheses open parentheses x close parentheses end cell equals cell 4 x squared minus 26 x plus 32 end cell end table 

Perhatikan bahwa f open parentheses 1 close parentheses diperoleh ketika g open parentheses x close parentheses equals 1. Sehingga dapat ditentukan x agar g open parentheses x close parentheses equals 1 sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses x close parentheses end cell equals 1 row cell negative x plus 3 end cell equals 1 row x equals cell 3 minus 1 end cell row x equals 2 end table 

Sehingga dapat ditentukan nilai f open parentheses 1 close parentheses sebagai berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses 1 close parentheses end cell equals cell f open parentheses g open parentheses x close parentheses equals 1 close parentheses end cell row blank equals cell f open parentheses g open parentheses 2 close parentheses close parentheses end cell row blank equals cell 4 times 2 squared minus 26 times 2 plus 32 end cell row blank equals cell 16 minus 52 plus 32 end cell row blank equals cell negative 4 end cell end table 

Jadi, jawaban yang benar adalah B.

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Diberikan fungsi f(x)=2x+5 dan g∘f(x)=2x+38x+21​,x=−23​, maka nilai g(3) adalah ...

Pembahasan Soal:

Dari soal diketahui 

f open parentheses x close parentheses equals 2 x plus 5

g ring operator f open parentheses x close parentheses equals fraction numerator 8 x plus 21 over denominator 2 x plus 3 end fraction comma x not equal to negative 3 over 2 comma

rumus g open parentheses x close parentheses dapat ditentukan dengan cara berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell g ring operator f open parentheses x close parentheses end cell equals cell g open parentheses f open parentheses x close parentheses close parentheses end cell row cell fraction numerator 8 x plus 21 over denominator 2 x plus 3 end fraction end cell equals cell g open parentheses 2 x plus 5 close parentheses end cell row blank blank blank end table 

misal a equals 2 x plus 5 left right double arrow x equals fraction numerator a minus 5 over denominator 2 end fraction 

table attributes columnalign right center left columnspacing 0px end attributes row cell fraction numerator 8 open parentheses begin display style fraction numerator a minus 5 over denominator 2 end fraction end style close parentheses plus 21 over denominator 2 open parentheses begin display style fraction numerator a minus 5 over denominator 2 end fraction end style close parentheses plus 3 end fraction end cell equals cell g open parentheses a close parentheses end cell row cell fraction numerator 4 open parentheses a minus 5 close parentheses plus 21 over denominator a minus 5 plus 3 end fraction end cell equals cell g open parentheses a close parentheses end cell row cell fraction numerator 4 a minus 20 plus 21 over denominator a minus 2 end fraction end cell equals cell g open parentheses a close parentheses end cell row cell fraction numerator 4 a plus 1 over denominator a minus 2 end fraction end cell equals cell g open parentheses a close parentheses end cell end table 

sehingga rumus g open parentheses x close parentheses equals fraction numerator 4 x plus 1 over denominator x minus 2 end fraction 

Nilai g open parentheses 3 close parentheses dapat ditentukan dengan cara berikut.

table attributes columnalign right center left columnspacing 0px end attributes row cell g open parentheses 3 close parentheses end cell equals cell fraction numerator 4 open parentheses 3 close parentheses plus 1 over denominator 3 minus 1 end fraction end cell row blank equals cell 13 over 2 end cell row blank equals cell 6 1 half end cell end table

 

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Roboguru

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