Is the inverse of a function #(x + 2)^2 - 4# a function?

1 Answer
Jul 23, 2015

It depends on the domain you pick. If #x# varies over the entire real line, the answer is "no". If #x# varies over a restricted domain, the answer can be "yes".

Explanation:

If you let #f# be the function, defined for all #x\in RR# by the formula #f(x)=(x+2)^2-4#, then this function does not pass the horizontal line test (lots of horizontal lines go through its graph, which is a parabola, more than once) and does not have an inverse function.

You could say, however, that it has an "inverse relation ". If you reflect the (full) graph of #y=f(x)=(x+2)^2-4# across the diagonal line #y=x#, you'll get a sideways parabola that does not pass the vertical line test (lots of vertical lines go through its graph more than once). Such a graph cannot be the graph of a function, which must have a unique output for each input. However, mathematicians still give a name to such an object. They call it a relation.

On the other hand, if you restrict the domain of the original function #f(x)=(x+2)^2-4# to, for instance, #x\geq -2#, then the graph is just the right half of the original parabola and passes the horizontal line test, and therefore has an inverse function. If you solve the equation #y=(x+2)^2-4# for #x \geq -2#, you'll get a formula for the inverse function with #y# as the independent variable: #x=f^{-1}(y)=-2+sqrt{y+4}#. If you decide to switch the variables around to write #y=f^{-1}(x)=-2+sqrt{x+4}# and graph this, you'll see the reflection property.

On the other hand once again, if you restrict the domain of the original function #f(x)=(x+2)^2-4# to, for instance, #x\leq -2#, then the graph is just the left half of the original parabola and passes the horizontal line test, and therefore has an inverse function. If you solve the equation #y=(x+2)^2-4# for #x \leq -2#, you'll get a formula for the inverse function with #y# as the independent variable: #x=f^{-1}(y)=-2-sqrt{y+4}#. If you decide to switch the variables around to write #y=f^{-1}(x)=-2-sqrt{x+4}# and graph this, you'll see the reflection property again (but for the other half of the original parabola).