What is the range of the function #f(x) = (3x-4)/(1+2x)#?

1 Answer
Jan 5, 2017

Answer:

The range is #=RR-{3/2}#

Explanation:

As you cannot divide by #0#, #1+2x!=0#, #=>#, #x!=-1/2#

The domain of #f(x)# is #D_f(x)=RR-{-1/2}#

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

There is a horizontal asymptote #y=3/2#

Therefore the range is #R_f(x)=RR-{3/2}#

graph{(y-(3x-4)/(1+2x))(y-3/2)=0 [-18.02, 18.01, -9.01, 9.01]}