How do you graph $f (x) = {(\frac{1}{3})}^{x + 1} + 2$ $f(x)=(1/3)^(x+1)+2$ and state the domain and range?

Question

1731 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-08T13:42:35

See below.

$f (x) = {(\frac{1}{3})}^{x + 1} + 2$ $f(x)=(1/3)^(x+1)+2$ is continuous, so there are no restrictions on $x$ $x$

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]}

How do you graph f(x)=(1/3)^(x+1)+2f(x)=(13)x+1+2f(x)=(1/3)^(x+1)+2 and state the domain and range?