Pertanyaan

F u n g s i f : R → d an g : R → R d i t e n t u kan o l e h f ( x ) = 3 x − 2 d an g ( x ) = x − 1 x u n t u k x  = 1 , maka ( f o g ) ( x ) a d a l ah ...

$F u n g s i f : R \to d an g : R \to R d i t e n t u kan o l e h f (x) = 3 x - 2 d an g (x) = \frac{x}{x - 1} u n t u k x \neq = 1, maka (f o g) (x) a d a l ah ...$

$fraction numerator 5 x minus 2 over denominator x minus 1 end fraction$
$fraction numerator 5 x plus 2 over denominator x minus 1 end fraction$
$fraction numerator x plus 1 over denominator x minus 1 end fraction$
$fraction numerator x minus 2 over denominator x minus 1 end fraction$
$fraction numerator x plus 2 over denominator x minus 1 end fraction$

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Hanya di Brain Academy

Habis dalam

Klaim

D. Kamilia

Master Teacher

Mahasiswa/Alumni Universitas Negeri Malang

Jawaban terverifikasi

Pembahasan

$left parenthesis f o g right parenthesis left parenthesis x right parenthesis space equals space f left parenthesis g left parenthesis x right parenthesis right parenthesis left parenthesis f o g right parenthesis left parenthesis x right parenthesis equals f open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses space space space space space space space space space space space space space equals 3 open parentheses fraction numerator x over denominator x minus 1 end fraction close parentheses minus 2 space space space space space space space space space space space space space equals fraction numerator 3 x over denominator x minus 1 end fraction minus 2 space space space space space space space space space space space space space equals fraction numerator 3 x minus 2 left parenthesis x minus 1 right parenthesis over denominator x minus 1 end fraction space space space space space space space space space space space space space equals fraction numerator x plus 2 over denominator x minus 1 end fraction$