How do you find the inverse of #f(x) = 4(x + 5)^2 - 6#?

Question

Serenity Johnson · Accepted Answer

No inverse function exists.

If there is no boundary on the domain, #f(x)# has no inverse function . It would have been possible, however, if the was a boundary like #x >= -5#. Why is this? Let me show you how to compute the inverse if there was one. First, set #y = f(x)#: #y = 4(x+5)^2 - 6# Next, exchange the roles of #x# and #y#: #x = 4(y+5)^2 - 6# Now, try to solve this equation for #y# in terms of #x#: ... add #6# on both sides ... # x + 6 = 4(y+5)^2# ... divide by #5# on both sides... # (x+6)/4 = (y+5)^2# Unfortunately there is no unique way to invert the square function since both the positive and the negative roots are solutions (E.g. #x^2 = 9# has two solutions: #x = 3# and #x = -3# since both #3^2 = 9# and #(-3)^2 = 9# hold.) This means that we would get something like: # +- sqrt((x+6)/4) = y+5# And finally, adding #-5# on both sides gives us: #y = -5 +- sqrt ((x+6)/4)# However, this means that for a unique value of #x# you don't get a unique value of #y# but two different values. This means that this is no function and thus, your problem has no solution.

Madelyn Anderson · Answer

undefined To find the inverse of $ f(x) = 4(x + 5)^2 - 6 $, follow these steps:

1. Replace $ f(x) $ with $ y $.
2. Swap $ x $ and $ y $, resulting in $ x = 4(y + 5)^2 - 6 $.
3. Solve the equation for $ y $ to find the inverse function.

After swapping $ x $ and $ y $, the equation becomes:

$$ x = 4(y + 5)^2 - 6 $$

Now, solve for $ y $:

$$ x = 4(y + 5)^2 - 6 $$
$$ x + 6 = 4(y + 5)^2 $$
$$ \frac{x + 6}{4} = (y + 5)^2 $$
$$ \pm \sqrt{\frac{x + 6}{4}} = y + 5 $$
$$ \pm \sqrt{\frac{x + 6}{4}} - 5 = y $$

So, the inverse function is:

$$ f^{-1}(x) = \pm \sqrt{\frac{x + 6}{4}} - 5 $$