How do you solve #(t-3)/(t+6)>0# using a sign chart?

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Answer 1 · 2018-02-25T12:47:06

Solution: # t < -6 and t > 3 # In interval notation:
#t| (-oo,-6)uu (3,oo)#

# (t-3)/(t+6)>0 ; t != -6# as the function is undefined at #t=-6#

Critical points are #t=3 and t = -6#

Sign chart:

When #t <-6 # sign of #f(t)# is #(-)/(-)=(+) ; >0#

When #-6 < t<3 # sign of #f(t)# is #(-)/(+)=(-) ; <0#

When #t > 3 # sign of #f(t)# is #(+)/(+)=(+) ; >0#

Solution: # t < -6 and t > 3 # . In interval notation:

#t| (-oo,-6)uu (3,oo)#