How do you graph #f(x)=(1/3)^(x+1)+2# and state the domain and range?

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Answer 1 · 2017-10-08T13:42:35

See below.

#f(x)=(1/3)^(x+1)+2# is continuous, so there are no restrictions on #x#

So domain is: #color(blue)({x in RR })#

#y# axis intercept where #x= 0#

#y=(1/3)^(0+1)+2 = color(red)(7/3)#
As:

#lim_(x->+oo)(1/3)^(x+1)=0#

So:

#lim_(x->+oo)(1/3)^(x+1)+2=2#

#color(red)(y=2)# is a horizontal asymptote.

For #x<-1#

#(1/3)^(x+1)=> 1/((1/3)^(x+1)#

#lim_(x->-oo)((1/3)^(x+1)+2=oo#

So range is:

#color(blue)({y in RR : 2< y < oo})#

Graph of #f(x)=(1/3)^(x+1)=2#:

graph{y= (1/3)^(x+1)+2 [-118.3, 118.8, -59.2, 59.4]}