domain and range of inverse functions?

Question

equation:

f(x) = (3x+4) / (2x-3)
D(-oo, 1.5)U(1.5, oo)
R(-oo, 1)U(1,oo)

f^-1(x) = (-3x-4) / (3-2x)
D(-oo, 1)U(1,oo)
R(-oo, 1.5)U(1.5, oo)

Is this correct? Thanks for your help.

Answer 1

Let f(x) = y

y = (3x+4)/(2x-3)

1) y(2x-3) = (3x+4)

2) 2xy-3y = 3x+4

3) 2xy-3x = 3y+4

4) x(2y-3) = (3y+4)

5) x = (3y+4)/(2y-3) "switch y to x"

6) y = (3x+4)/(2x-3) "switch y to f^-1(x)"

7) f^-1(x) = (3x+4)/(2x-3)

* from f(x) = (3x+4)/(2x-3) "2x-3 = 2x = 3 = x can't = 3/2"
Domain = (-infinity,3/2)u(3/2,infinity)

* from f^-1(x) = (3x +4)/(2x-3) "2x-3 = 2x = 3 x can't = 3/2"
Range = (-ininity,3/2)u(3/2,infinity)

Answer 2

I think that the range for the first equation is not quite correct. y can equal 1 when x= -7. the y value will never approach 1.5. So the the range is actually the same as the domain.

For the inverse, the range is the domain of the original function and the domain of the inverse is the range of the original. You had the right idea but you defined teh range of the original incorrectly so you just need to make that change. That means that the inverse of this function is really the same as the original.

For evidence of this all you have to look at is that when rearranged, the inverse function's equation is the same as the original's. By taking the negative from the top and moving it to the bottom you get the original function.

(-3x-4) / (3-2x) = -(3x+4)/(3-2x) = (3x+4)/-(3-2x) = (3x+4)/(-3+2x) = (3x+4)/(2x-3)

Answer 3

It looks like you've got your domains and ranges close but I disagree with your f^-1(x); that's not right.

An inverse function is defined as one whose value is thus:

y = f(x)
x=f^-1(y)

So, the domain of the inverse function is the range of the original function and the range of the inverse function is the domain of the original function.

in the case of f(x) = (3x+4)/(2x-3),

the domain is indeed (-inf,1.5) union (1.5,+inf), i.e. every number except +1.5, where f(x)-> -inf as x-> 1.5 from below and f(x) -> +inf as x->1.5 from above. By the way, that means the range of f^-1(x) is (-inf,1.5) union (1.5,+inf), i.e. every number except +1.5 .

The range of f(x) looks to me to be all reals except 1.5, not all reals except 1. The reason I say that is that the limit as x->+inf would seem to be 3/2 = 1.5, and the limit as x-> -inf also equals 1.5 . The domain and range, to me seem to be the same.

So I'm having trouble understanding how you determined that f(x) could never equal 1, not saying you're wrong, just saying that as I envision the graph, it would seem to permit the value to be 1.

I hope I'm not steering you wrong.

Answer 4

As Christopher showed, your range for f(x) is wrong.

Take the limit of f(x) as x--> 1.5.
You'll see that f(x) approaches 1.5 not 1

Answer 5

permit f(x) = arccos(x - a million) - 2. -a million ? x - a million ? a million 0 ? x ? 2 -?/2 ? arccos(x - a million) ? ?/2 -?/2 - 2 ? arccos(x - a million) - 2 ? ?/2 - 2 area of f(x): [0, 2] selection of f(x): [-?/2 - 2, ?/2 - 2] permit y = f?¹(x). f[f?¹(x)] = x f(y) = x arccos(y - a million) - 2 = x arccos(y - a million) = x + 2 y - a million = cos(x + 2) y = cos(x + 2) + a million f?¹(x) = cos(x + 2) + a million area of f?¹(x): [-?/2 - 2, ?/2 - 2] selection of f?¹(x): [0, 2] particular, cos(x + 2) + a million would be evaluated on any genuine quantity, yet an x outdoors of the style of f(x) can't be on the area of f?¹(x).

Answer 6

(-3x-4)/(3-2x) is identical to (3x+4)/(2x-3), not its inverse
Let y=f(x)=(3x+4)/(2x-3)
then
2xy-3y = 3x+4
x(2y-3)=3y-4
x=(3y-4)/(2y-3)