help me to answer this question.. given f(x)=5/2-x for x<2. find f inverse and its domain..

Let's replace f(x) with y.... \[\large y = \frac52-x\]is this correct?

yes.. but i don't know their domain...

can u help me find the domain?

after a few more steps it will become apparent you want the inverse of this equation right? take @terenzreignz equation...and switch the x's and y's

can u show me? bcoz it have range for x<2.. so i'm not sure..

More on that later, for now, just switch the x and y, and show me what you got.

y=5/2-x

exactly.....since the inverse...is the same a the original equation...we can conclude this has no inverse correct?

Excuse me? The inverse of this function is not the original equation... ;)

oops....lol wow such a rookie mistake...forgive my last post

go back to finding the inverse...we have x = 5/2 - y correct? solve for y again

Or it is... I'm not really sure :D

turns out it is... forgive MY last posts :D

Just seemed too good to be true.

wait really?

Yeah.. inverse of a-x is also a-x, for any constant a. a-(a-x) = x

x = 5/2 - y 2x = 5 - 2y -2y = 2x - 5 y = -x + 5/2 ahh looky there lol

Okay, @sha0403 To get the inverse of a function, switch x and y, and attempt to express y as purely an expression of x.

Step by step method courtesy of @johnweldon1993

lol, yeah sorry for the "spoonfeeding"...but I had to confirm with myself

but okay...since that is the inverse (original equation) we want to find it's domain correct?

but, what the function of x<2 in the question? given?

Yes :) okay, back to \[\huge f(x)=\frac52-x\] And its domain is all x < 2 right?

yes

Now, if we consider its inverse (which, either annoyingly or conveniently, is itself) we get \[\huge f^{-1}(x)=\frac52-x\] What do we know about inverses and the domain of the original function?

The range of the inverse must be the domain of the original function, right?

right...so can u explain the domain of the function in more specific, i'm not really understand it

The domain of a function f is all values of x for which f(x) has a value. Take for instance \[\huge g(x)=\frac1x\] Its domain is all real numbers EXCEPT 0, since g does not have a value at g(0)

and for x<2? what the function of that? just ignore it or what?

No. This is what you call restricting the domain. If left without the x<2 Then we have \[\huge f(x)=\frac52-x\]and its domain is all real numbers because, well, let's face it, no matter what real number you have for x, the function will be defined. But the domain CAN be restricted to a smaller set. In this case, the domain could have been all real numbers, but instead, the domain is RESTRICTED to x<2 or \[\large x\in(-\infty,2)\]

thank u very much.. i get it now..=)

So... how about the inverse of the function? That the range of the inverse of the function is the domain of the original function?

i not sure,,

Okay :) Let y = f(x) that is to say, y is a function of x. The domain of f(x) is the set of all PERMISSIBLE values of x The range of f(x) is the set of all POSSIBLE values of y, given that x comes from the domain. An example... \[\huge y =h(x)=|x|\] the absolute value function... any real number can be evaluated under h (the absolute value) so its domain is... \[\huge dom \ h =\mathbb{R}\]

for the domain,, why the 2 is include? while the given x<2?

2 is not included :)

ok2 tq sir..

Now, the range of h... h is the absolute value, so it can only ever churn out non-negative numbers so... \[\huge ran \ h = [0,\infty)\]

Back to your question... \[\huge f^{-1}(x)=\frac52-x\]

We're asked for the domain, and we're given the domain of the original function, f(x) which is x < 2, right?

oh yes..SO?

Well, the range of the inverse is the domain of the original function, is that clear to you now? :)

so the domain for inverse function is negative infinity to 2?

No... that is the domain for the original function :) The RANGE for the inverse function is negative infinity to 2. You following me so far? Don't worry, we're almost done ;)

i ask u how to find the domain of inverse function right? or have misunderstanding at there?

We are getting there. Just tell me if you understand everything so far.

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