The notation of a circle between two functions (for example, *f* o *g* (*x*)) means the composition of two functions. It is found as function f (the one that comes first) of *g*(*x*):

f o *g (*x*)*= f(*g*(x)).

In this case, the two functions in question are the function *f* and its inverse, denoted as *f*^(-1). (Note that the exponent of -1 here means inverse, NOT the reciprocal, which would be 1/f.)

Since the inverse of a function is created by swapping the independent and dependent variables, it takes the value of *f* as the independent variable and converts it to original input variable x, making it dependent variable. If the function and its inverse are composed, the result is the original variable, *x*:

*f* o *f ^*(-1) *(*x*)*= *f*(*f*^(-1)(*x*)) = *x*.

In other words, the composition of a function and its inverse (or vice versa) is the identity function: it equals *x.*

For the given function *y* = *x*/4 + 3, this can be confirmed by obtaining the inverse function explicitly and then finding the composition.

The inverse function is obtained by exchanging *x* and *y,* and then solving for *y*:

*x* = *y*/4 + 3

*x* - 3 = *y*/4

*y* = 4(*x* - 3)

Now, if the original function *f*(*x*) =* x*/4 + 3 is evaluated at 4(*x*-3) instead of *x*, we get

*f* o *f*^(-1) (*x*) = [4(*x* - 3)]/4 + 3 = *x* - 3 + 3 = *x*, as expected.

fof^-1(x)

= f(f^-1(x))

= x, by definition

Here y = x/4 + 3

The inverse function is found by expressing y in terms of x, x = (y - 3)*4. Now interchange y and x. f^-(x) = (x - 3)*4

fof^-1(x)

= f((x - 3)*4)

= (x - 3)*4/4 + 3

= x

**This proves that ****for f(x) = x/4 + 3, ****fof^-1(x) = x **