How do you solve and write the following in interval notation: #1+ 20/(x + 1) ≤ 20/x#?

1 Answer
May 10, 2017

Answer:

The solution is #x in [-5,1) uu (0,4]#

Explanation:

We cannot do crossing over.

So, we rewrite, simplify and factorise the inequality

#1+(20)/(x+1)<=20/x#

#(x+1+20)/(x+1)<=20/x#

#(x+1+20)/(x+1)-20/x<=0#

#(x(x+21)-20(x+1))/(x(x+1))<=0#

#(x^2+21x-20x-20)/(x(x+1))<=0#

#(x^2+x-20)/(x(x+1))<=0#

#((x-4)(x+5))/(x(x+1))<=0#

Let #f(x)=((x-4)(x+5))/(x(x+1))#

We can construct a sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##-5##color(white)(aaaaa)##-1##color(white)(aaaaaa)##0##color(white)(aaaaaa)##4##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x+5##color(white)(aaaaaa)##-##color(white)(aaaa)##+##color(white)(aa)##||##color(white)(aa)##+##color(white)(aa)##||##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x+1##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aa)##+##color(white)(aa)##||##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x##color(white)(aaaaaaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aa)##-##color(white)(aa)##||##color(white)(aaa)##+##color(white)(aaa)##+#

#color(white)(aaaa)##x-4##color(white)(aaaaaa)##-##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aa)##-##color(white)(aa)##||##color(white)(aaa)##-##color(white)(aaa)##+#

#color(white)(aaaa)##f(x)##color(white)(aaaaaaa)##+##color(white)(aaaa)##-##color(white)(aa)##||##color(white)(aa)##+##color(white)(aa)##||##color(white)(aaa)##-##color(white)(aaa)##+#

Therefore,

#f(x)<=0# when #x in [-5,1) uu (0,4]#