How do you solve #((x-1)(3-x))/(x-2)^2>=0#?

2 Answers
Nov 30, 2016

Answer:

# f(x)≥0# for #1≤x≤3# with the exclusion of #x=2# where it is undefined.

Explanation:

The denominator is always positive, so the sign of #f(x)# is determined by the numerator.

The numerator is positive when both factors are positive or both are negative.

#(x-1) >= 0# for #x>=1#
#(3-x) >=0# for #x<=3#

So #f(x) >=0# for #1<=x<=3# with the exclusion of #x=2# where it is undefined.

Nov 30, 2016

Answer:

The answer is #x in [ 1,2[ uu ]2, 3 ] #

Explanation:

Let #f(x)=((x-1)(3-x))/(x-2)^2#

The domain of #f(x)# is #D_f(x)=RR-{2}#

Also the denominator is #>0#, #AAx in D_f(x)#

So, we make the sign chart

#color(white)(aaaa)##x##color(white)(aaaa)##-oo##color(white)(aaaa)##1##color(white)(aaaa)##3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##x-1##color(white)(aaaa)##-##color(white)(aaa)##+##color(white)(aaaa)##+#

#color(white)(aaaa)##3-x##color(white)(aaaa)##+##color(white)(aaa)##+##color(white)(aaaa)##-#

#color(white)(aaaa)##f(x)##color(white)(aaaaa)##-##color(white)(aaa)##+##color(white)(aaaa)##-#

So,

#f(x)>=0# when #x in [ 1,2[ uu ]2, 3 ] #

graph{((x-1)(3-x))/(x-2)^2 [-12.66, 12.65, -6.33, 6.33]}