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

Question

Christian Antunez · Accepted Answer

No inverse function exists without domain restrictions.

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. ======================= 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)#

Elias Ackerman · Answer

undefined To find the inverse of $ f(x) = (x + 2)^2 - 4 $, you need to switch the roles of $ x $ and $ y $ and then solve for $ y $.

1. Start with the given function:
   $$ f(x) = (x + 2)^2 - 4 $$

2. Replace $ f(x) $ with $ y $:
   $$ y = (x + 2)^2 - 4 $$

3. Swap $ x $ and $ y $:
   $$ x = (y + 2)^2 - 4 $$

4. Solve for $ y $:
   $$ x = (y + 2)^2 - 4 $$
   $$ x + 4 = (y + 2)^2 $$
   $$ \sqrt{x + 4} = y + 2 $$
   $$ y = \sqrt{x + 4} - 2 $$

So, the inverse of $ f(x) = (x + 2)^2 - 4 $ is $ f^{-1}(x) = \sqrt{x + 4} - 2 $.