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What can be said about fof^-1(x) of the function:  y = x/4 + 3.

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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.

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= 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


= f((x - 3)*4)

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

= x

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

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