inverse of this function

[m1ke]

alright, I need a bit of help finding the inverse function of f(x)=-2(x+1)²+3

I keep getting: f¹(x)= [(x-1)/2]² but i know this is wrong

[The Girl From Ipanema]

this might be totally wrong because I haven't taken math in ages but could it be....


f¹(x)= (x-3)/-2 <<<< with this whole thing being square rooted?????

[m1ke]

yep, I think the teacher picked an equation that is impossible to be used because there are restrictions. and yeah you replace y with f(x) and then switch y with x

f(x)=-2(x+1)²+3
y=-2(x+1)²+3
-2(y+1)²+3=x
-2(y+1)²=x-3
-2(y+1)²= x-3
-2-2
y= x-3-1

-2

ah you get it, I know I screwed up with the 2 and a 1 but it still doesnt work.. all of my friends are stuck on it also. Im sure that it is an impossible equation that the teacher made just to make him feel smart for tomorrow. Im going to kick him in the shin if it is. oh well thanks to anyone that tried to get it

[The Girl From Ipanema]

Originally Posted by m1ke

yep, I think the teacher picked an equation that is impossible to be used because there are restrictions. and yeah you replace y with f(x) and then switch y with x

f(x)=-2(x+1)²+3
y=-2(x+1)²+3
-2(y+1)²+3=x
-2(y+1)²=x-3
-2(y+1)²= x-3
-2-2
y= x-3-1

-2

ah you get it, I know I screwed up with the 2 and a 1 but it still doesnt work.. all of my friends are stuck on it also. Im sure that it is an impossible equation that the teacher made just to make him feel smart for tomorrow. Im going to kick him in the shin if it is. oh well thanks to anyone that tried to get it

Did you read what i posted above you? cause I think you made that past at the same time as me making mine

[m1ke]

Originally Posted by The Girl From Ipanema

Did you read what i posted above you? cause I think you made that past at the same time as me making mine

lol, I dont know thats why I think my teacher is trying to be a smartass.

my friend is say that he got:


f¹(x)= -1 +/- [(x-5)/-2]

the [] being square root.

[The Girl From Ipanema]

Originally Posted by m1ke

lol, I dont know thats why I think my teacher is trying to be a smartass.

my friend is say that he got:


f¹(x)= -1 +/- [(x-5)/-2]

the [] being square root.

that looks really off to me .....but I coudln't really say for sure sorry

[That Kid with the Hair]

The answer is [x-3/-2] - 1 = f-1(x).

The inverse of a given function "f1()" is annotated as "f-1()" not "f2()" most commonly by mathematicians.

[King Bowzer]

The inverse of a function is just a function that maps f(x) to x; and the inverse of the inverse would map x back to f(x). I.e., the inverse of the inverse of f(x) is just f(x).

Given a function f such that f(x) = -2(x+1)² + 3 is the same thing as saying this you're mapping x to -2(x+1)² + 3. But you want a function f such that

f(-2(x+1)² + 3) = x; or equivalently, a function that maps -2(x+1)² + 3 to x.

In simpler terms, your original function gives you -2(x+1)² + 3 when you plug in a value for x. The inverse of that function will give you x when you plug in -2(x+1)² + 3.

To find the inverse, just swap f(x) for x in the equation and solve for f(x).

x = -2(f(x) + 1)² + 3

x - 3 = -2(f(x) + 1)²

(-1/2)(x - 3) = (f(x) + 1)²

sqrt[(-1/2)(x - 3)] = f(x) + 1

fi(x) = sqrt[(-1/2)(x - 3)] - 1

This is the inverse of your original function (as denoted by the i). You can verify this by seeing if the inverse function maps the original function back to x; which it should.

fi(f(x)) = sqrt[(-1/2)((-2(x+1)² + 3) - 3)] - 1

= sqrt[(-1/2)(-2(x+1)²)] - 1

= sqrt[(x+1)²] - 1

= x+1 - 1

= x

There you have it.