The function f is defined by fx=5/(1-3x) for x>1 Find an expression for f^-1 and state domain and range of f^-1?

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Answer 1 · 2017-11-19T07:34:02

See the explanation below

Let #y=5/(1-3x)#, #AA x>1#

Therefore,

#y(1-3x)=5#

#y-3xy=5#

#3xy=y-5#

#x=(y-5)/(3y)#

So,

#f^-1(x)=(x-5)/(3x)#

The domain of #f^-1(x)# is #RR-{0}#

and the range is #RR-{1/3}#

If #x>1#, then the domain is #(1,+oo)#

and

the range is

#((5/(1-3)), 0)#, #=>#, #(-2.5,0)#

See the curve

graph{5/(1-3x) [-10, 10, -5, 5]}