Pertanyaan

Diketahui fungsi f : R → R ditentukan dengan rumus f ( x ) = 2 x + 1 x + 3 ; x  ≡ − 2 1 . Invers fungsi f(x) adalah f -1 (x) = ...

Diketahui fungsi $f : R \to R$ ditentukan dengan rumus $f (x) = \frac{x + 3}{2 x + 1}; x \neq \equiv - \frac{1}{2}$ . Invers fungsi f(x) adalah f^-1(x) = ...

$fraction numerator negative x plus 3 over denominator 2 x minus 1 end fraction semicolon x not identical to 1 half$
$fraction numerator negative x minus 3 over denominator 2 x minus 1 end fraction semicolon x not identical to 1 half$
$fraction numerator negative x minus 3 over denominator 2 x plus 1 end fraction semicolon x not identical to negative 1 half$
$fraction numerator negative x plus 3 over denominator 2 x plus 1 end fraction semicolon x not identical to negative 1 half$
$fraction numerator x plus 3 over denominator 2 x plus 1 end fraction semicolon x not identical to negative 1 half$

S. Intan

Master Teacher

Mahasiswa/Alumni Institut Pertanian Bogor

Jawaban terverifikasi

Pembahasan

Perhatikan letak a dan d ditukar dan tandanya berubah jadi negatif (-) , maka

$f left parenthesis x right parenthesis equals fraction numerator a x plus b over denominator c x plus d end fraction rightwards arrow f to the power of negative 1 end exponent open parentheses x close parentheses equals fraction numerator negative d x plus b over denominator c x plus open parentheses negative a close parentheses end fraction$

Perhatikan letak a dan d ditukar dan tandanya berubah jadi negatif (-) , maka

$f left parenthesis x right parenthesis equals fraction numerator x plus 3 over denominator 2 x plus 1 end fraction rightwards arrow f to the power of negative 1 end exponent left parenthesis x right parenthesis equals fraction numerator negative x plus 3 over denominator 2 x minus 1 end fraction$

Fungsi Komposisi

Fungsi Invers

Invers Fungsi Komposisi

Latihan Soal Fungsi Komposisi dan Invers

Latihan Bab