What is the inverse of #y=3ln(5x)+x^3# ?

1 Answer
May 14, 2016

Answer:

#f^(-1)(y) = x : f(x) = y#

Explanation:

Let #f(x) = 3ln(5x)+x^3#

Let us assume that we are dealing with Real values and therefore the Real natural logarithm.

Then we are constrained to #x > 0# in order that #ln(5x)# be defined.

For any #x > 0# both terms are well defined and so #f(x)# is a well defined function with domain #(0, oo)#.

Note that #3ln(5)# and #x^3# are both strictly monotonic increasing on this domain so our function is too and is one-to-one.

For small positive values of #x#, the term #x^3# is small and positive and the term #3ln(5x)# is arbitrarily large and negative.

For large positive values of #x#, the term #3ln(5x)# is positive and the term #x^3# is arbitrarily large and positive.

Since the function is also continuous, the range is #(-oo, oo)#

So for any value of #y in (-oo, oo)# there is a unique value of #x in (0, oo)# such that #f(x) = y#.

This defines our inverse function:

#f^(-1)(y) = x : f(x) = y#

That is #f^(-1)(y)# is the value of #x# such that #f(x) = y#.

We have shown (informally) that this exists, but there is no algebraic solution for #x# in terms of #y#.

The graph of #f^(-1)(y)# is the graph of #f(x)# reflected in the line #y=x#.

In set notation:

#f = { (x, y) in (0, oo) xx RR : y = 3ln(5x)+x^3 }#

#f^(-1) = { (x, y) in RR xx (0, oo) : x = 3ln(5y)+y^3 }#