How do I find if this function is a one to one, and is that the same as finding the inverse?

Question

first off, is finding the one to one the same as finding the inverse, and if so, how do I find if this is a one to one.

f(x)=(sqrt x-2)/(x-3)

some detail explanation would be great, because I am a little lost.

Answer 1

Note: I'm assuming that you mean f(x) = (√x - 2)/(x-3). If you mean f(x) = √(x-2)/(x-3), this solution won't be applicable to your problem.

To show a function is one-to-one is to show that if f(x)=f(y), then x=y, or equivalently that if x≠y, then f(x)≠f(y). It is certainly sufficient to find an inverse function (since ∃f⁻¹ s.t. f⁻¹(f(x)) = x means that f(x)=f(y) ⇒ f⁻¹(f(x)) = f⁻¹(f(y)) ⇒ x=y), but it is not required -- anything that shows f(x)=f(y) ⇒ x=y will work. Conversely, to show that the function is not 1-1, it suffices to show that there are two distinct numbers x and y such that f(x)=f(y), but x≠y. In this case, a quick look at the graph of the function tells us we will probably be able to find a nonzero x such that f(x)=f(0). So let us compute f(0):

f(0) = -2/-3 = 2/3

So now let us solve the equation f(x) = 2/3, and see if we can get a solution other than 0:

(√x - 2)/(x-3) = 2/3
3√x - 6 = 2x - 6
2x = 3√x (we're looking for a nonzero solution, so √x≠0, so:
2√x = 3
√x = 3/2
x=9/4

And let us check this:

f(9/4) = (√(9/4) - 2)/(9/4-3) = (3/2 - 2)/(-3/4) = (-1/2)/(-3/4) = 2/3 = f(0) ✓

Since f(9/4) = f(0), but 9/4≠0, this function is not one to one.

Edit: of course in graduate level mathematics you'd be correct SM, but at this student's level the difference between codomain and range still isn't stressed, so their textbook probably defines the inverse of a function f as a function from the range (as opposed to codomain) of f to the domain of f such that f⁻¹(f(x)) = x (I remember that mine did). So I'm just trying to match the set of definitions that they're likely to be working with.

It's interesting to note that if the domain of f⁻¹ is the range of f (either because you're using the definitions from a calc 1 book or because f is known to be surjective), then f⁻¹(f(x)) = x for all x in the domain actually does imply that f(f⁻¹(x)) = x for all x in the range. Why? Because if x∈range (f), then x=f(y) for some y in the domain, so f(f⁻¹(x)) = f(f⁻¹(f(y))) = f(y) = x. So you often can get away with just checking one condition.

Answer 2

Finding one to one is not the same as finding the inverse, in your own words. I think what you meant to say was that if we know that the function is one to one, does that mean that the function has an inverse. The answer is NO. Just because a function is one to one, does NOT mean that it is also invertible. In order to be invertible, a function must be one to one AND onto.

Now, the second question, how do you determine if a function is one to one or not. Use the definition, a function f(x) is said to be one to one if f(a)=f(b) implies that a=b. So just do that and see what happened.

Your f(x)=(sqrt(x)-2)/(x-3).

So starting with the fact that
(sqrt(a)-2)/(a-3)=(sqrt(b)-2)/(b-3)
see if you can simplify this into a=b. If you can, then f is one to one, if you can't, then f is not one to one.


Edit
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Good answer Pascal and probably easier to understand but one slight mistake, in order to make sure that the inverse is the inverse function you have to verify BOTH
f(f^-1(x))=x
AND
f^-1(f(x))=x

Just checking one way is not enough.

Answer 3

that's kinda an identical ingredient you probably did on your occasion, even nonetheless that's amped somewhat. here we bypass... y= (x-2)/(x+2) clean up for x. Multiply the two facets by ability of (x + 2). y(x + 2) = (x-2)/(x+2) * (x + 2). y(x + 2) = x - 2 Distribute the y. yx + 2y = x - 2 Subtract 2y from the two facets. yx + 2y - 2y = x - 2 -2y yx = x - 2 -2y Subtract x from the two facets. yx - x = x - 2 -2y - x yx - x = - 2 -2y ingredient out the x from the two words on the left. x(y - a million) = - 2 -2y Divide the two facets by ability of (y - a million). x(y - a million) / (y - a million) = (- 2 -2y) / (y - a million) x = (- 2 -2y) / (y - a million) Now you swith x and y. y = (- 2 -2x) / (x - a million) you may get equivalent types for this, however the respond is unquestionably what you spot above.