The function F(x) = (x − 3)^2 − 4 for x ∈ (−∞, 3] is one-to-one. What is its inverse function? What is the domain and range of F^ −1 ?

Answer 1

F(x) = (x-3)^2 - 4

F(x) = y

y = (x-3)^2 -4

switch x and y and solve for the new y

x = (y-3)^2 -4

(y-3)^2 = x +4

y-3 =+ or - sqr(x+4)

y = + or - sqr(x+4) + 3

F^-1 =+ or - sqr(x+4) + 3

It's not an inverse function though. It's an inverse relation, but not a function since there's 2 y values for one x value. It doesn't meet the vertical line test for functions. there is no inverse function, just an inverse relation.

However, since the domain of F(x) is x< or =3 then F^-1 <or =3 and F^-1 = -sqr(x+4)+3 is an inverse function. While + or - sqr(x+4)+3 is not a function. You need to use just the negative square root.

the domain of the inverse function is x greater than or equal to -4, since x<-4 would make the sqr(x+4) an imaginary number

F(x) is an upward opening parabola with vertex or minimum point (3,-4) but just the left side of the parabola given the domain has x < or equal to -3

F^-1(x) is a rightward opening parabola but with just the bottom half of the parabola since y is always < or =3