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Pertidaksamaan

Contoh 1
Tentukan interval dan Hp serta gambarkan garis bilangan dari penyelesaian pertidaksamaan : $-2 < \frac{x+5}{2}< 3$

Penyelesaian :
$-2 < \frac{x+5}{2}< 3$ (semua ruas dikalikan 2)
$-2.2< x+5< 3.2$
$-4< x+5< 6$ (semua ruas dikurangi 5)
$-4-5< x< 6-5$
$-9< x< 1$
Interval : $\left ( -9,1 \right )$
$Hp = \left \{ x|-9< x< 1,x\in R \right \}$

Contoh 2
Himpunan penyelesaian pertidaksamaan : $\frac{x^{2}-5x+4}{x^{2}+2x-3} \geqslant 0$

Penyelesaian:
$\frac{x^{2}-5x+4}{x^{2}+2x-3} \geqslant 0$
$\frac{\left ( x-1 \right )\left ( x-4 \right )}{\left ( x-1 \right )\left ( x+3 \right )}\geqslant 0$
$\frac{\left ( x-4 \right )}{\left ( x+3 \right )}\geqslant 0$
$\left ( x-4 \right )\left ( x+3 \right )= 0$
$x-4= 0$ atau $x+3= 0$
$x= 4$ atau $x= -3$
$Hp=\left \{ x|x< -3 \cup x> 4, x \in R\right \}$

Contoh 3
Himpunan penyelesaian pertidaksamaan : $\left | 2x-4\right |< 6$

Penyelesaian:
$\left | 2x-4\right |< 6$
$-6< 2x-4< 6$
$-6+4< 2x< 6+4$
$-2< 2x< 10$
$\frac{-2}{2}< x<\frac{10}{2}$
$-1< x<5$
$Hp=\left \{ x| -1< x< 5,x\in R\right \}$

Fungsi

Contoh 1:
Diketahui : $f\left ( x \right )=7x-5$ dan $g\left ( x \right )=2x^{2}+x$ . Tentukan :
$a. \left ( fog \right )\left ( x \right )$
$b. \left ( gof \right )\left ( x \right )$

Penyelesaian:
$a. \left ( fog \right )\left ( x \right )=f\left ( g\left (x\right ) \right )$
$=f\left ( 2x^{2}+x \right )$
$=7\left ( 2x^{2}+x \right )-5$
$=14x^{2}+7x-5$

$b. \left ( gof \right )\left ( x \right )=g\left ( f\left ( x \right ) \right )$
$=g\left ( 7x-5 \right )$
$=2\left ( 7x-5 \right )^{2}+\left ( 7x-5 \right )$
$=2\left ( 49x^{2}-70x+25 \right )+7x-5$
$=98x^{2}-140x+50+7x-5$
$=98x^{2}-133x+45$

Contoh 2:
Diketahui : $f\left ( x \right )=\frac{2x+1}{3-x}$ dan $g\left ( x \right )= \sqrt{x^{2}-1}$ . Tentukan :
$a. \left ( fog \right )\left ( x \right )$
$b. \left ( gof \right )\left ( x \right )$

Penyelesaian:
$a. \left ( fog \right )\left ( x \right )=f\left ( g\left ( x \right ) \right )$
$=f\left (\sqrt{x^{2}-1}\right )$
$=\frac{2\sqrt{x^{2}-1}+1}{3-\sqrt{x^{2}-1}}$

$b. \left ( gof \right )\left ( x \right )=g\left ( f\left ( x \right ) \right )$
$=g\left ( \frac{2x+1}{3-x} \right )$
$= \sqrt{\left ( \frac{2x+1}{3-x} \right )^{2}}-1$

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