How do you solve and write the following in interval notation: #x / (x-9) = 0#?

Question

How do you solve and write the following in interval notation: #x / (x-9) >= 0#?

Eva Adamson · Accepted Answer

The answer is # x in ] -oo, 0] uu ] 9,+ oo[ #

Let #f(x)=x/(x-9)#

The domain of #f(x)# is #D_f(x)=RR-{9}#

Let's do a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##0##color(white)(aaaa)####color(white)(aaaa)##9##color(white)(aaaaa)##+oo#

#color(white)(aaaa)##x##color(white)(aaaaaaa)##-##color(white)(aaaaaa)##+##color(white)(a)####color(white)(a)##∥##color(white)(aa)##+#

#color(white)(aaaa)##x-9##color(white)(aaaa)##-##color(white)(aaaaaa)##-##color(white)(a)####color(white)(a)##∥##color(white)(aa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##+##color(white)(aaaaaa)##-##color(white)(a)####color(white)(a)##∥##color(white)(aa)##+#

Therefore,

#f(x)>=0# when # x in ] -oo, 0] uu ] 9,+ oo[ #

Jameson Allen · Answer

undefined To solve $ \frac{x}{x-9} \geq 0 $ and write it in interval notation, we need to find the critical points where the expression equals zero or is undefined, and then determine the intervals where the expression is positive or zero.

The critical points occur where the numerator is zero (i.e., $ x = 0 $) and where the denominator is zero (i.e., $ x = 9 $).

We divide the real number line into intervals using these critical points: $ (-\infty, 0), (0, 9), $ and $ (9, \infty) $.

Testing each interval with a test point, we find that $ \frac{x}{x-9} \geq 0 $ holds true for the intervals $ (-\infty, 0) $ and $ (9, \infty) $.

Thus, the solution to the inequality in interval notation is $ (-\infty, 0] \cup (9, \infty) $.

How do you solve and write the following in interval notation: #x / (x-9) &gt;= 0#?

How do you solve and write the following in interval notation: #x / (x-9) >= 0#?