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For f: R -> R defined by

3x - 5, x > 0;

f(x) = {

-3x + 1, x <= 0,

determine f^-1([-5,5]).

Can someone help out and show how to do this. This is a sample problem for a test.

3x - 5, x > 0;

f(x) = {

-3x + 1, x <= 0,

determine f^-1([-5,5]).

Can someone help out and show how to do this. This is a sample problem for a test.

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f(x) is one-to-one and linear over x>0 and over x<=0 separately, so the domain that gives the range -5,5 should be easy to find. Then you just set y=f(x) and solve for x (again separately for the different pieces).

If so, do you just solve for x, and then plug in the range [-5, 5] into the y values?

Example 1: y = 3x -5

--> x = (y + 5)/3

--> x = (5 + 5)/3

--> x = 10/3

Example 2: y = -3x + 1

--> x = (y - 1)/-3

--> x = (-5 - 1)/-3

--> x = 2

Still a bit confused as you can tell.

For x>0: x = (y + 5)/3

For x<=0: x = (y - 1)/-3

But you need to find the ranges separately for each of the two options.

So for x>0

(-5 + 5)/3 = 0

(5 + 5)/3 = 10/3

So the range over [-5,5] is [0,10/3]

But x > 0 so the range of valid answers should be (0, infinity) so the real range is (0, 10/3].

For x<=0

(-5 - 1)/-3 = 2

(5 - 1)/-3 = -4/3

So the range over [-5, 5] is [-4/3, 2]

But x <= 0, so the range of valid answers is (-infinity, 0] so the real range is [-4/3, 0]

I'm not sure how the answer is supposed to look, but I would write it something like this

(y + 5)/3 for a<y<b

f^-1(y) = { undefined for c<y<d

(y - 1)/-3 for g<y<h

Some of those '<'s should be '<=' probably.

I'll leave it to you do try to determine the domains. But the typical notion of a function is that it cannot be multivalued so in the cases where you can't tell which f(x) was used, I would call it undefined.

Does that make sense.

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Start your 7-day free trial"But you need to find the ranges separately for each of the two options."

if you have set A and B where a is a member of A and b is a member of B, and if f(a)=b, then f^-1(b) = a, right?

According to above statement, b is in the range, and a is in the domain.

Then -->according to f^-1([-5,5]) --> [-5,5] is the range right, and the inverse would be the domain right? Then how come your quote above says that you need to find the RANGES separately. Don't we already know the Range to be [-5,5], and we need to find the DOMAIN not the RANGE?

In your case, the domain of f is A and the range of f is B but for f^-1 it's flipped. The domain of f^-1 is B and the range is A.

The function f: Z --> N defined by

2x - 1, if x > 0;

f(x) = {

-2x, otherwise.

is one-to-one and onto. Fine f^-1.

If f is Z->N, it doesn't really change anything. Just f^-1 will of course be N->Z

The function f: Z --> N defined by

2x - 1, if x > 0;

f(x) = {

-2x, otherwise.

is one-to-one and onto. Fine f^-1.

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