How do you solve #2/(p-1)>=3/4# using a sign chart?

4 Answers
Jan 6, 2018

Answer:

#p in (1,11/3]#

Explanation:

#2/(p-1)>=3/4#

#2/(p-1)-3/4>=0#

#(2*4-3(p-1))/(4(p-1))>=0#

#(11-3p)/(4(p-1))>=0#

zero points:
#p_1=11/3#

#p_2=1#

#[[,|,-oo;1,1;11/3,11/3;oo],[11-3p,|,+,+,-],[p-1,|, -,+,+]]#

#p in (-oo,1)hArr# function is negative

#color(red)(p in (1,11/3)hArr "function is positive")#

#p in (11/3,oo)hArr# function is negative

But the function may equal to zero which is only when #p=11/3#. Therefore we include 11/3 to the asnwer.

Jan 6, 2018

Answer:

The solution is #p in (1,11/3]#

Explanation:

Simplify the inequality

#2/(p-1)>=3/4#

#2/(p-1)-3/4>=0#

Putting on the same denominator

#((2*4-3*(p-1)))/(4(p-1))>=0#

#(8-3p+3)/(4(p-1))>=0#

#(11-3p)/(4(p-1))>=0#

Let #f(p)=(11-3p)/(4(p-1))#

Construct the sign chart

#color(white)(aaaa)##p##color(white)(aaaa)##-oo##color(white)(aaaaaa)##1##color(white)(aaaaaaa)##11/3##color(white)(aaaa)##+oo#

#color(white)(aaaa)##p-1##color(white)(aaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aaaaa)##+#

#color(white)(aaaa)##11-3p##color(white)(aaa)##+##color(white)(aaaa)####color(white)(aaaaa)##+##color(white)(aa)##0##color(white)(aa)##-#

#color(white)(aaaa)##f(p)##color(white)(aaaaaa)##-##color(white)(aaaa)##||##color(white)(aaaa)##+##color(white)(aa)##0##color(white)(aa)##-#

Therefore,

#f(p) >=0# when #p in (1,11/3]#

graph{(11-3x)/(4(x-1)) [-12.66, 12.65, -6.33, 6.33]}

Jan 6, 2018

Answer:

I never really did them this way. But, here is an excellent tutorial and example: http://www.fmaths.com/signcharts/lesson.php

Explanation:

Check the link to see the steps required for this particular problem.

Jan 6, 2018

Answer:

Ulternate approach

Explanation: