How do you find the inverse of #f(x) =(x + 2)^2 - 4#?

1 Answer
Jan 25, 2016

No inverse function exists without domain restrictions.

Explanation:

Set #f(x) = y#:

#y = (x+2)^2 - 4#

Interchange #y# and #x# in your equation:

#x = (y+2)^2 - 4#

Now, you need to solve this equation for #y#.
First of all, add #4# on both sides:

#x + 4 = (y+2)^2#

The next step would be to draw the root. However, this will leave you with two solutions, since e.g. for #25 = x^2#, both #5 = x# and #-5 = x# are solutions.

#sqrt(x+4) = abs(y+2)#

#<=> +-sqrt(x + 4) = y + 2#

Subtract #2# on both sides:

# -2 +- sqrt(x+4) = y#

Beware that a function must have a unique value for #y# for each unique value of #x#.

However, this is not the case here since for e.g. #x = 12#, you have both #y = - 2 + sqrt(16) = 2# and #y = - 2 - sqrt(16) = -6#.

This means that there no inverse function exists.

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Remark:

An inverse function would exist if you restricted the domain of the original function.

As you can easily see that the vertex of the function is at #x = -2#, it would suffice to either restrain the domain to e.g. #x <= -2# or to #x >=-2#.

For example, if your original function was

#f(x) = (x+2)^2 -4 " where " x >=-2#

then you could continue with the calculation from above, abandoning the negative term:

#y = -2 + sqrt(x+4)#

Replace #y# with #f^(-1)(x)#:

#f^(-1)(x) = -2 + sqrt(x+4)#