Finding and graphing inverse functions question  TOPIC_SOLVED

Complex numbers, rational functions, logarithms, sequences and series, matrix operations, etc.

Finding and graphing inverse functions question

Postby Hwy61Meg on Mon Mar 18, 2013 12:12 am

Ok, I have a test tomorrow, and I've gotten myself really confused here. I'm working on exercises from my book, and the answers are in the back.

The problem:
f(x) = (x-1)2 , x is < or = 1.

Find the inverse and graph both f and f-1

Now, I followed the steps to find the inverse thusly:

1. Change f(x) to y: y = (x-1)2 , x is < or = 1.
2. Interchange x and y: x = (y-1)2 , y is < or = 1.
3. Solve for y: I ended up with y = the square root of x, + 1 (+ 1 outside the radical).
Here is the problem. The book says the answer is: f-1 = NEGATIVE the square root of x, +1. I'm completely confused about how that ended up being negative.

Any tips?
Hwy61Meg
 
Posts: 2
Joined: Mon Mar 18, 2013 12:02 am

Sponsor

Sponsor
 

  TOPIC_SOLVED

Postby stapel_eliz on Mon Mar 18, 2013 12:51 am

Hwy61Meg wrote:OThe book says the answer is: f-1 = NEGATIVE the square root of x, +1. I'm completely confused about how that ended up being negative.

The restricted domain ("x < 1") is what tells which "half" of the function is in play, and thus which of the two roots (because square-rooting gave you a "plus-minus") should be used for the inverse's domain. You can see an example in the last exercise on this page.

Because the original function used only the left-hand "half" of the parabola, the inverse function must (after flipping over the line y = x) use only the bottom "half" of the sideways parabola, and thus only the negative root.

Hope this helps! :wink:
User avatar
stapel_eliz
 
Posts: 1708
Joined: Mon Dec 08, 2008 4:22 pm

Re: Finding and graphing inverse functions question

Postby Hwy61Meg on Mon Mar 18, 2013 1:19 am

It does, thank you. The only other restriction in my example problems is a positive restriction, so it wasn't obvious that they'd "chosen" the positive radical. I had forgotten about that +/- when taking the square root. They haven't been using that notation in the problems for this section, and our instructor hasn't been using them when solving for inverses. But it makes sense now, with that restriction.

So, the graphed line for f(x) in this case is half the graph of a parabola (due to the restriction), which starts, after the transformation, at (1,0), and the inverse is the graph of a square root function, which starts, after its transformation, at (0,1) and falls below the x axis? I have a picture of the graph in my book but I can't share it. It was confusing to me because the x and y intersections aren't marked with a dot and the two lines overlap, so it looks like one continuous curve with no end points, if that makes sense. So I was having trouble reverse-engineering my problem to try and figure out what I was doing.
Hwy61Meg
 
Posts: 2
Joined: Mon Mar 18, 2013 12:02 am


Return to Advanced Algebra ("pre-calculus")

cron