Fungsi Komposisi

Contoh 1 

Diketahui$latex f(x)=3x+5$dan$latex g(x)=7-3x$. Tentukan pemetaan$latex x=2$oleh fungsi$latex f(x)$dilanjutkan$latex g(x)$. 

Jawaban: 
Alternatif 1

$latex f(2)=3.2+5=6+5=11\\g(11)=7-3.11=7-33=-26$

Jadi peta$latex x=2$oleh fungsi$latex f(x)$dilanjutkan$latex g(x)$adalah $latex -26$.

Alternatif 2

$latex \begin{array}{rcl}(g\circ f)(x)&=&g\big(f(x)\big)\\&=&g(3x+5)\\&=&7-3(3x+5)\\&=&7-9x-15\\&=&-8-9x\end{array}$

maka 

$latex (g\circ f)(2)=-8-9.2=-8-18=-26$
Jadi peta$latex x=2$oleh fungsi$latex f(x)$dilanjutkan$latex g(x)$adalah $latex -26$.

Contoh 2 

Diketahui$latex f(x)=x+7$dan$latex (f\circ g)(x)=5x+3$. Tentukan$latex g(x)$. 

Jawaban: 

$latex \begin{array}{rcl}(f\circ g)(x)&=&f\big(g(x)\big)\\5x+3&=&g(x)+7\\5x+3-7&=&g(x)\\5x-4&=&g(x)\end{array}$

Jadi$latex g(x)=5x-4$.

Contoh 3 

Jika$latex g(x-5)=7x+3$, tentukan$latex g(x)$. 

Jawaban: 
Alternatif 1

$latex \begin{array}{rcl}g(x-5)&=&7x+3\\g(x-5)&=&7(x-5)+35+3\\g(x-5)&=&7(x-5)+38\\g(x)&=&7x+38\end{array}$

Alternatif 2

Misalkan$latex x-5=y\qquad \Rightarrow \qquad x= y+5$

maka

$latex \begin{array}{rcl}g(x-5)&=&7x+3\\g(y)&=&7(y+5)+3\\g(y)&=&7y+35+3\\g(y)&=&7y+38\\g(x)&=&7x+38\end{array}$

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Contoh 4 

Diketahui$latex (g\circ f)(x)=2x^2+5x-5$dan$latex f(x)=x-1$, maka$latex g(x)=....$ 

Jawaban: 

$latex \begin{array}{rcl}(g\circ f)(x)&=&2x^2+5x-5\\g\big(f(x)\big)&=&2x^2+5x-5\\g(x-1)&=&2x^2+5x-5\end{array}$

Alternatif 1

$latex \begin{array}{rcl}g(x-1)&=&2(x-1)^2+4x-2+5x-5\\&=&2(x-1)^2+9x-7\\&=&2(x-1)^2+9(x-1)+2\\g(x)&=&2x^2+9x+2\end{array}$

Alternatif 2

Misalkan$latex y=x-1\qquad \Rightarrow \qquad x= y+1$

maka

$latex \begin{array}{rcl}g(y)&=&2(y+1)^2+5(y+1)-5\\&=&2(y^2+2y+1)+5y+5-5\\&=&2y^2+4y+2+5y\\&=&2y^2+9y+2\\g(x)&=&2x^2+9x+2\end{array}$

Contoh 5 

Diketahu$latex (g\circ f)(x)=4x^2+4x$dan$latex g(x)=x^2-1$dan berlaku$latex f(x)>1$untuk$latex x>-\frac{1}{2}$, maka$latex f(x-2)=....$ 

Jawaban: 

$latex \begin{array}{rcl}(g\circ f)(x)&=&4x^2+4x\\g\big(f(x)\big)&=&4x^2+4x\\\big(f(x)\big)^2-1&=&4x^2+4x\\\big(f(x)\big)^2&=&4x^2+4x+1\\\big(f(x)\big)^2&=&(2x+1)^2\end{array}$

Karena$latex f(x)>1$untuk$latex x>-\frac{1}{2}$ maka diperoleh 

$latex \begin{array}{rcl}f(x)&=&2x+1\\f(x-2)&=&2(x-2)+1\\f(x-2)&=&2x-4+1\\f(x-2)&=&2x-3\end{array}$

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