Pertanyaan

Diketahui f ( x ) = 2 x + 3 , g ( x ) = 3 x − 1 , dan ( h ∘ g ∘ f ) ( x ) = x + 4 x 2 − 1 . Tentukan nilai h ( 0 ) .

Diketahui $f (x) = 2 x + 3$ , $g (x) = 3 x - 1$ , dan $(h \circ g \circ f) (x) = \frac{x ^{2} - 1}{x + 4} .$ Tentukan nilai $h (0) .$

A. Salim

Master Teacher

Mahasiswa/Alumni Universitas Pelita Harapan

Jawaban terverifikasi

Jawaban

didapat nilai dari h ( 0 ) = 96 28 = 24 7

didapat nilai dari

h (0) = \frac{28}{96} = \frac{7}{24}

Pembahasan

Ingat bahwa ( h ∘ g ∘ f ) ( x ) = h ( g ( f ( x ))) . Sehingga ( h ∘ g ∘ f ) ( x ) h ( g ( f ( x ))) h ( g ( 2 x + 3 ) h ( 3 ( 2 x + 3 ) − 1 ) h ( 6 x + 9 − 1 ) h ( 6 x + 8 ) = = = = = = x + 4 x 2 − 1 x + 4 x 2 − 1 x + 4 x 2 − 1 x + 4 x 2 − 1 x + 4 x 2 − 1 x + 4 x 2 − 1 Misalkan: 6 x + 8 = y maka 6 x x = = y − 8 6 y − 8 Jadi, h ( 6 x + 8 ) h ( y ) h ( y ) h ( x ) = = = = = = = = x + 4 x 2 − 1 ( 6 y − 8 ) + 4 ( 6 y − 8 ) 2 − 1 ( 6 y − 8 ) + 4 ( 36 y 2 − 16 y + 64 ) − 1 6 y − 8 + 6 24 36 y 2 − 16 y + 64 − 36 36 6 y + 16 36 y 2 − 16 y + 28 36 6 ( y + 16 ) 36 y 2 − 16 y + 28 6 ( y + 16 ) y 2 − 16 y + 28 6 ( x + 16 ) x 2 − 16 x + 28 Sehingga nilai dari h ( 0 ) adalah h ( x ) h ( 0 ) = = = = 6 ( x + 16 ) x 2 − 16 x + 28 6 ( 0 + 16 ) ( 0 ) 2 − 16 ( 0 ) + 28 6 ( 16 ) 0 − 0 + 28 96 28 Dengan demikian, didapat nilai dari h ( 0 ) = 96 28 = 24 7

Ingat bahwa $(h \circ g \circ f) (x) = h (g (f (x)))$ . Sehingga

$(h \circ g \circ f) (x) h (g (f (x))) h (g (2 x + 3) h (3 (2 x + 3) - 1) h (6 x + 9 - 1) h (6 x + 8) = = = = = = \frac{x ^{2} - 1}{x + 4} \frac{x ^{2} - 1}{x + 4} \frac{x ^{2} - 1}{x + 4} \frac{x ^{2} - 1}{x + 4} \frac{x ^{2} - 1}{x + 4} \frac{x ^{2} - 1}{x + 4}$

Misalkan: $6 x + 8 = y$ maka

$6 x x = = y - 8 \frac{y - 8}{6}$

Jadi,

$h (6 x + 8) h (y) h (y) h (x) = = = = = = = = \frac{x ^{2} - 1}{x + 4} \frac{( \frac{y - 8}{6} ) ^{2} - 1}{( \frac{y - 8}{6} ) + 4} \frac{( \frac{y ^{2} - 16 y + 64}{36} ) - 1}{( \frac{y - 8}{6} ) + 4} \frac{\frac{y ^{2} - 16 y + 64}{36} - \frac{36}{36}}{\frac{y - 8}{6} + \frac{24}{6}} \frac{\frac{y ^{2} - 16 y + 28}{36}}{\frac{y + 16}{6}} \frac{\frac{y ^{2} - 16 y + 28}{36}}{\frac{6 ( y + 16 )}{36}} \frac{y ^{2} - 16 y + 28}{6 ( y + 16 )} \frac{x ^{2} - 16 x + 28}{6 ( x + 16 )}$

Sehingga nilai dari $h (0)$ adalah

$h (x) h (0) = = = = \frac{x ^{2} - 16 x + 28}{6 ( x + 16 )} \frac{( 0 ) ^{2} - 16 ( 0 ) + 28}{6 ( 0 + 16 )} \frac{0 - 0 + 28}{6 ( 16 )} \frac{28}{96}$