Set f(x)=f and solve for x...

Here is ihavelovedyousincewewere18's  original question for part e.....

e. Set f(x)=4 and solve for x. Show on your graph that one solution is valid and the other one is extraneous. (Plot 2 points showing one is on the graph and the other is not.)

4 = x - √[x- 2]      rearrange

√[x- 2]  = x - 4      square both sides

 x - 2  = x^2 - 8x + 16       subtract x  and add 2 to both sides

x^2 - 9x + 18  = 0     factor

(x - 6) (x - 3)  = 0      and setting both factors to 0, we have that x = 6 or  x = 3

Notice that x = 6 is a valid answer to the original problem  because 6 - √[6- 2]  = 6 - √4  =  6 - 2  = 4

But x = 3  isn't valid because  6 - √[3- 2]  = 6 - √1  =  6 - 1  = 5

Here's the graph........https://www.desmos.com/calculator/uakphwshii

Notice that (6, 4) is on the graph, but (3, 5)  isn't

I think there's something missing here, ihavelovedyousincewewere18

CPhill, that is so sweet:) I had no idea you had a sweetheart on here. <3

what's missing, CPhill?? Thanks for helping btw :)

I don't understand  ...  f(x)  = f       what is "f"  (???)