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?

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?>>

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.

I did not mean x<-2, but x is less than or equal to. How is this going to change the answer?

Explain Bot Explain Bot answered
9 months ago

Apologies for the confusion earlier. If we have f(x) = -(x-2)^2, x <= 2 (with x less than or equal to 2), we can still find the inverse in a similar way. Here's the corrected process:

1. Start with the given equation: y = -(x-2)^2.
2. Swap x and y: x = -(y-2)^2.
3. Solve for y: x = -(y-2)^2.
4. Take the square root of both sides: sqrt(x) = -(y-2).
5. Multiply both sides by -1: -sqrt(x) = y-2.
6. Add 2 to both sides: -sqrt(x) + 2 = y.
7. Replace y with f^(-1)(x) (the inverse function notation): f^(-1)(x)= -sqrt(x) + 2.

Therefore, the inverse function of f(x) = -(x-2)^2, x <= 2 is f^(-1)(x) = -sqrt(x) + 2.

The fact that x is less than or equal to 2 does not affect finding the inverse function. The steps remain the same. However, it's important to note the domain of the inverse function will be restricted to the range of the original function. In this case, the domain of f^(-1)(x) = -sqrt(x) + 2 will be x greater than or equal to 0.