อ้างอิง:
ข้อความเดิมเขียนโดยคุณ Siren-Of-Step
I have just studied relations : Domain and Range Mr. Metree Sritongtae
Find Domain and Range
3.$ r = \left\{\,\right. (x,y) \in R \times R \left|\,\right. y = \frac{1}{x^2-2x-3} \left.\,\right\} $
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From$x^2-2x-3 \not= 0$
$(x-3)(x+1) \not= 0$
$x \not= 3,-1$
$D_r = R-(\left\{\,\right. 3\left.\,\right\} \cup \left\{\,\right. -1\left.\,\right\} )$
หา $R_r$
$\frac{1+3y}{y}= x^2-2x$
$\frac{1+3y}{y}+1 = (x-1)^2$
$\frac{1+4y}{y} \geqslant 0$
$y(4y+1) \geqslant 0$ $y \not= 0$
$(-\infty ,\frac{-1}{4}] \cup (0,\infty )$