my question was:
find f^-1 (x). (this is asking me to find the inverse)
f(x) = -(x-2)^2, x <= 2
how do I solve this problem?>>
and you answered:
If f(x)=-(x-2)^2 x<-2, then
let y=f(x)
- y = (x-2)^2
sqrt(-y)= x-2
x(y)= sqrt (-y) + 2
g(x)= sqrt(-x) + 2
and that is the inverse function
check:
g(f(x))= sqrt( (x-2)^2) +2 =x
f(g(x)= -((sqrt(-x) + 2 -2)^2= x
since g(f(x))=f(g(x)=x, then g(x) above is the inverse. Practice these.
However, I didn't mean x<-2, but x<=2. Is this going to make a difference in the final answer?
yes, you can see that taking the root of -x means x is <=0, therefore you need to restrict the domain.
Incidentally, there's a relatively simple way to "see" what the inverse of a function looks like. If you're,respond.
could you tell me more about it?
This is a very simple device and you may or may not have seen it.
Follow the steps below
1. You need square piece of paper, but clear plastic like an overhead transparency is better. It should be square though.
2.From the lower left hand corner to the upper right hand corner, draw the diagonal. This is the line y = x, but it's also called the identity function f(x) = x.
3. Put the x and y axis through the middles of the paper so the origin is in the center.
4. Now you can draw an arbitrary function on the paper.
5. Now you flip the paper over along that diagonal.
6. Now you should be able to see the axis's and the function on the other side. Label the axis on the flipped side so that was the x axis in the neg. direction is now the pos. y axis on the flip side; the - y axis pre-flip is now + x axis post-flip.
7. The function you see through the paper post-flip is the inverse of the function pre-flip.
What you have done is reflect the function across the identity function. This is one way to view an inverse function. Try this with the function f(x) = x^2 and you will see why we only use the pos. branch of root-x. You should also see why 1-1 functions have inverses; I don't know if you've covered these.
As I said, this demo works best with clear plastic.
I should also mention that this is sometimes an exercise when you study something called the dihedral group D4; the group of rotations and reflections of a group of order 4. I don't expect you to understand this, but there is some group theory going on here too.
Yes, the fact that x <= 2 instead of x <-2 will make a difference in the final answer for the inverse function.
To find the inverse function, let's first rewrite the given function with the correct inequality:
f(x) = -(x-2)^2, x <= 2
Now, let y = f(x):
y = -(x-2)^2
To find the inverse, we need to solve for x in terms of y.
1. Start by multiplying both sides by -1:
-y = (x-2)^2
2. Take the square root of both sides, being careful with the negative sign:
sqrt(-y) = x - 2
3. Finally, add 2 to both sides to isolate x:
x = sqrt(-y) + 2
Now, we have x in terms of y:
x(y) = sqrt(-y) + 2
Therefore, the inverse function of f(x) = -(x-2)^2, x <= 2 is:
g(x) = sqrt(-x) + 2
Now, let's check if g(x) is indeed the inverse function by confirming if g(f(x)) = x and f(g(x)) = x.
1. g(f(x)):
g(f(x)) = g(-(x-2)^2)
= sqrt(-(-(x-2)^2)) + 2
= sqrt((x-2)^2) + 2
= |x-2| + 2
Since the given inequality is x <= 2, we need to take the positive square root when calculating the inverse function. Therefore, we have:
g(f(x)) = x - 2 + 2
= x
2. f(g(x)):
f(g(x)) = -((sqrt(-x) + 2) - 2)^2
= -sqrt(-x)^2
= -(-x)
= x
Since we also have f(g(x)) = x, we can conclude that g(x) = sqrt(-x) + 2 is indeed the inverse function of f(x) = -(x-2)^2, x <= 2.
I hope this clarifies the solution. Let me know if you have any further questions!