What is the inverse function of #f(x)= absx + 1#?

2 Answers
dani83
Aug 16, 2015

No inverse as #f(x)# is not a one-to-one function. However, it is possible to obtain a multivalued inverse.

Multivalued inverse:
# f^(-1)(x) = +-(x-1), " " x >= 1#

Explanation:

Multivalued inverse:
# f(x) = |x| + 1 #
# f(x)-1 = |x| #
# f^(-1)(x) = +-(x-1), " " x >= 1#

# f^(-1)(x) # can also be found by taking y = x as the axis of symmetry.
Note that the range of f(x) becomes the domain of #f^(-1)(x)#.

George C.
Aug 16, 2015

As a function from #RR# to #RR#, #f(x)# is not one-one.

As a result it has no well defined inverse function.

Explanation:

For example, #f(1) = f(-1) = 2#, so #f^(-1)(2)# is not well defined - it could be #1# or #-1#.

If we restrict the domain of #f(x)# to #[0, oo)# or to #(-oo, 0]#, then the resulting mapping is one-one onto #[1, oo)# and there is a well defined inverse function with domain #[1, oo)#.

#f_{[0, oo)}(x)# has inverse #f_{[0, oo)}^-1(y) = y - 1# where #y in [1, oo)#

#f_{(-oo, 0]}(x)# has inverse #f_{(-oo, 0]}^-1(y) = 1 - y# where #y in [1, oo)#

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