Jika f(x+2)=4x+12x−3 dan g(x)=5−x24x+1, nilai dari (2f−3g)(2) adalah ....

Pertanyaan

Jika $begin mathsize 14px style f open parentheses x plus 2 close parentheses equals fraction numerator 2 x minus 3 over denominator 4 x plus 1 end fraction end style$ dan $begin mathsize 14px style g open parentheses x close parentheses equals square root of fraction numerator 4 x plus 1 over denominator 5 minus x squared end fraction end root end style$ , nilai dari $\style{font-size:14px}{\left(2f-3g\right)\left(2\right)}$ adalah ....

I. Kumaralalita

Master Teacher

Mahasiswa/Alumni Universitas Gadjah Mada

Jawaban terverifikasi

Jawaban

nilai

adalah

Pembahasan

Pertama tentukan fungsi terlebih dahulu, misalkan maka sehingga:

$begin mathsize 14px style table attributes columnalign right center left columnspacing 0px end attributes row cell f open parentheses x plus 2 close parentheses end cell equals cell fraction numerator 2 x minus 3 over denominator 4 x plus 1 end fraction end cell row cell f open parentheses y close parentheses end cell equals cell fraction numerator 2 open parentheses y minus 2 close parentheses minus 3 over denominator 4 open parentheses y minus 2 close parentheses plus 1 end fraction end cell row blank equals cell fraction numerator 2 y minus 4 minus 3 over denominator 4 y minus 8 plus 1 end fraction end cell row cell f open parentheses y close parentheses end cell equals cell fraction numerator 2 y minus 7 over denominator 4 y minus 7 end fraction end cell row cell f open parentheses x close parentheses end cell equals cell fraction numerator 2 x minus 7 over denominator 4 x minus 7 end fraction end cell end table end style$

Kemudian gunakan konsep nilai dari selisih dua fungsi:

maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 f minus 3 g close parentheses open parentheses x close parentheses end cell equals cell 2 f open parentheses x close parentheses minus 3 g open parentheses x close parentheses end cell row blank equals cell 2 open parentheses fraction numerator 2 x minus 7 over denominator 4 x minus 7 end fraction close parentheses minus 3 open parentheses square root of fraction numerator 4 x plus 1 over denominator 5 minus x squared end fraction end root close parentheses end cell end table$

Substitusikan x equals 2 ke dalam hasil selisih fungsi di atas, maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell open parentheses 2 f minus 3 g close parentheses open parentheses 2 close parentheses end cell equals cell 2 open parentheses fraction numerator 2 open parentheses 2 close parentheses minus 7 over denominator 4 open parentheses 2 close parentheses minus 7 end fraction close parentheses minus 3 open parentheses square root of fraction numerator 4 open parentheses 2 close parentheses plus 1 over denominator 5 minus open parentheses 2 close parentheses squared end fraction end root close parentheses end cell row blank equals cell 2 open parentheses fraction numerator 4 minus 7 over denominator 8 minus 7 end fraction close parentheses minus 3 open parentheses square root of fraction numerator 8 plus 1 over denominator 5 minus 4 end fraction end root close parentheses end cell row blank equals cell 2 open parentheses fraction numerator negative 3 over denominator 1 end fraction close parentheses minus 3 open parentheses square root of 9 over 1 end root close parentheses end cell row blank equals cell 2 open parentheses negative 3 close parentheses minus 3 open parentheses square root of 9 close parentheses end cell row blank equals cell negative 6 minus 3 open parentheses 3 close parentheses end cell row blank equals cell negative 6 minus 9 end cell row blank equals cell negative 15 end cell end table$

Dengan demikian, nilai adalah .