How do you find the inverse of #f(x)=x^2+2x-5# and is it a function?

Question

How do you find the inverse of  #f(x)=x^2+2x-5# and is it a function?

Serenity Jones · Accepted Answer

#f^1# does not exist.

Let us recall that, the inverse of a given function exists, if and only if , the given function is a bijection , i.e., the given function is a 1 - 1 ( one-to-one , or, injection ) and onto ( or, surjection ). Observe that, for the given function #f#, it is not injective, because, #f(x)=x^2+2x-5=x(x+2)-5#, so that, #f(0)=-5=f(-2)#, meaning that #f# is an #m-1# or, a #many-to-one# and not a #1-1# function. Hence, #f^1# does not exist.

Caleb Alexander · Answer

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

1. Replace $ f(x) $ with $ y $.
2. Swap the variables $ x $ and $ y $ to form an equation in terms of $ x $ and $ y $.
3. Solve the equation obtained in step 2 for $ y $, which will give you the expression for the inverse function $ f^{-1}(x) $.
4. Replace $ y $ with $ f^{-1}(x) $ to represent the inverse function.

Step 1: Replace $ f(x) $ with $ y $:
$ y = x^2 + 2x - 5 $

Step 2: Swap the variables $ x $ and $ y $:
$ x = y^2 + 2y - 5 $

Step 3: Solve for $ y $:
$ x = y^2 + 2y - 5 $
$ 0 = y^2 + 2y - x - 5 $

Using the quadratic formula, $ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}} $, where $ a = 1 $, $ b = 2 $, and $ c = -x - 5 $, we get:

$ y = \frac{{-2 \pm \sqrt{{2^2 - 4(1)(-x - 5)}}}}{{2(1)}} $
$ y = \frac{{-2 \pm \sqrt{{4 + 4x + 20}}}}{{2}} $
$ y = \frac{{-2 \pm \sqrt{{4x + 24}}}}{{2}} $
$ y = \frac{{-2 \pm \sqrt{{4(x + 6)}}}}{{2}} $
$ y = \frac{{-2 \pm 2\sqrt{{x + 6}}}}{{2}} $
$ y = -1 \pm \sqrt{{x + 6}} $

Step 4: Replace $ y $ with $ f^{-1}(x) $:
$ f^{-1}(x) = -1 \pm \sqrt{{x + 6}} $

Therefore, the inverse function of $ f(x) = x^2 + 2x - 5 $ is $ f^{-1}(x) = -1 \pm \sqrt{{x + 6}} $.

As for whether the inverse function is a function itself, it is not a function because it does not pass the vertical line test. For any given value of $ x $, there are two possible corresponding values of $ y $. Hence, it is not a one-to-one function and thus not a function.