Algebra Know-It-ALL

220 Mappings, Relations, and Functions

Solution

The inverse of g is not a function. Here’s the function again. Remember that the domain is the entire set

of reals, and the range is the set of nonnegative reals:

w=g(v)=v^2

If you take the equation w=v^2 and transpose the positions of the independent and dependent variables,

you get

v=w^2

This is the same as

w=±(v1/2)

The plus-or-minus symbol is important here! It indicates that for every value of the independent variable

v you plug into this relation, you’ll get two values of w, one positive and the other negative. You can also

write

g−^1 (v)=±(v1/2)

The function g is two-to-one (except when v= 0), and that’s okay. But the inverse is one-to-two (except

whenw= 0). That makes g−^1 a legitimate relation, but prevents it from being a function.

The other two functions above have inverses that are also functions. Both f and h are one-to-one, so

their inverses must also be one-to-one. We have

f(x)=x+ 1

and

f –^1 (x)=x− 1

We also have

h(t)=t^3

and

h−^1 (t)=t1/3

If a function is one-to-one over a certain domain and range, then you can transpose the values of the inde-

pendent and dependent variables while leaving their names the same, and you can transpose the domain

and range. The resulting inverse is a function. If a function is many-to-one, then its inverse is one-to-

many, and is therefore not a function.