What is the domain and range of #f(x) = 2 - e ^ (x / 2)#?

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Answer 1 · 2018-02-21T11:33:05

Domain: #(-oo,oo)#

Range: #(-oo,2)#

The domain is all possible values of #x# with which #f(x)# is defined.

Here, any value of #x# will result in a defined function. Therefore, the domain is #-oo<##x<##oo#, or, in interval notation:

#(-oo,oo)#.

The range is all possible values of #f(x)#. It can also be defined as the domain of #f^-1(x)#.

So to find #f^-1(x):#

#y=2-e^(x/2)#

Interchange the variables #x# and #y#:

#x=2-e^(y/2)#

And solve for #y#:

#x-2=-e^(y/2)#

#e^(y/2)=2-x#

Take the natural logarithm of both sides:

#ln(e^(y/2))=ln(2-x)#

#y/2ln(e)=ln(2-x)#

As #ln(e)=1#,

#y/2=ln(2-x)#

#y=2ln(2-x)=f^-1(x)#

We must find the domain of the above.

For any #lnx,# #x>0#.

So here, #2-x>0#

#-x> -2#

#x##<##2#

So the range of #f(x)# can be stated as #(-oo,2)#