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C3 Range/Domain

i=.jpg

The only thing left troubling me in C3 is Range/Domain.
How would one work out the range/domain of a function?

Take the attachment as an example.
(edited 8 years ago)
Reply 2


Great, thanks.
Reply 3
Examine what happens to the function at the outer points of the domain (be they plus or minus infinity or set values for x) and if necessary differentiate to find maxima/minima as well. For functions such as these looking at the asymptote is also probably necessary, although it just so happens that 2 is already the lower limit of the function (although x does not ever equal 2 of course) in this case so you will do that anyway. Drawing a graph can also be of great assistance.

For fractional functions like this, when you're looking at x tending to infinity, a simple trick is to get rid of the constants - a crude way of understanding this would be to say that infinity + c is still infinity (thinking of infinity is a number is never pleasant but oh well). So in this case you could get rid of the -3 and -2 to find out what the function is approaching as x tends to infinity. Also look at what is happening when x tends to 2, and in this case (as drawing a graph will confirm) there are no turning points to consider, so looking at the domain boundaries alone should guide you to the answer.
Reply 4
Original post by edothero
i=.jpg

The only thing left troubling me in C3 is Range/Domain.
How would one work out the range/domain of a function?

Take the attachment as an example.


Domain of a function:

The domain of a function is basically all the values that the function can take in (NB: this is really maximal domain -- why questions are so stupid at A-Level, I don't know...). Different functions can take in different values.

Basically, for a function f(x)\displaystyle f(x), the domain is all the xx-values that ff can take/are valid. Let's consider a few examples, that'll clear it up.

Square root function: f(x)=2x+3\displaystyle f(x) = \sqrt{2x + 3}. Now, you know that you can take the square root of pretty much any number as long as it is positive. Plug 2 into your calculator and square it, all is fine. Plug pi in there, still good. Plug in 100, it'll give you 10. 0 works. But as soon as you do -1, you're dead. -pi. Dead. -10. Dead. So you can see that the square root function wants whatever is under the .\sqrt{.} to be positive.

That is, for any value of x that you can possibly think of, you want to have 2x+32x+3 being positive, since that's what's under the square root. So your domain is 2x+30    2x3    x322x+3 \geq 0 \iff 2x \geq -3 \iff x \geq -\frac{3}{2}. So the domain is x32x \geq -\frac{3}{2}.

Spoiler


Rational function: f(x)=x+1x+2\displaystyle f(x) = \frac{x+1}{x+2}.

You know that for these kinds of function, any value of x you can think of will work. Except for one single value. One teensy little number that if you plug it will give you UNDEFINED on your calculator. Can you guess what it is? Yups! x=2x = -2 f**ks the function up.

So your domain is every value of x except 2. Or written mathematically, xR,x2x \in \mathbb{R} , x \neq 2.

Spoiler


Other functions have other requirements, for example, ln(x)ln(x) can take in only positive numbers and not 0. So x >0. What about e^x?

Range of a function:

The range of a function is the part of the y-axis that the function takes up. That is, the interval between the maximum value of the function and the minimum value of the function.

The graph y=x goes all the way from -infinity to infinity.

The exponential function: f(x)=exf(x) = e^x, this has domain xRx\in \mathbb{R}, or more simply, you can put any number into this function and it'll work. Negative, 0, Positive, whatevs.

Buuuut, whatever number you put in, the function is going to spit out a number that HAS TO BE POSITIVE. You can see this from the graph. e^(-1000), it'll give you a small positive number. The function goes all the way up to +infinity. But it'll never go below 0 or even touch 0.

So the range of the function is 0<ex<+0 < e^x < +\infty

Ranges and domain of inverse functions:

The range of an inverse function f1(x)f^{-1}(x) is the domain of the original function f(x)f(x).

The domain of an inverse function f1(x)f^{-1}(x) is the range of the original function f(x)f(x).

Anything else you need clearing up? Anything you didn't understand?
(edited 6 years ago)
Reply 5
Original post by edothero
i=.jpg

The only thing left troubling me in C3 is Range/Domain.
How would one work out the range/domain of a function?

Take the attachment as an example.


A simple way to do this in some cases is to write x in terms of f (or y, whatever you want to call the range), then use the inequality for the domain given to you in the question. However i just gave your example question a go using this method and it didn't quite work as y's ended up cancelling and you just get a mess! Unfortunately that happens sometimes. So i would differentiate to find maxima and minima. (You then have y<maximum or y>minimum.
you usually have to give asymptotes as well. to do this, for the domain, it can't equal anything that would mean you're dividing by zero (so if you have something over (2-x) then x cannot equal 2) and same for range (rewrite your equation with x in terms of y)
If you have the luxury of a graphic calculator, a sneaky method is to just have your calculator draw the graph for you, and you can just see any max/min points and asymptotes.
Reply 6
Original post by Zacken
x


This is gold. Thank you very much.

So with the function in the example I posted;

2x3x2 \dfrac{2x-3}{x-2}

The domain would be xR,x2x \in \mathbb{R} , x \neq 2 as xx can be any value but 2 because we would get 10 \dfrac{1}{0}

As for the range, we get something like this. I'm a bit confused with how to go about this as it seems the function covers y from -infinity to infinity; though the mark scheme says
"x>2f(x)>2x>2 \therefore f(x)>2"

FpxGeaf4.jpg
Reply 7
Original post by edothero
This is gold. Thank you very much.

So with the function in the example I posted;

2x3x2 \dfrac{2x-3}{x-2}

The domain would be xR,x2x \in \mathbb{R} , x \neq 2 as xx can be any value but 2 because we would get 10 \dfrac{1}{0}

As for the range, we get something like this. I'm a bit confused with how to go about this as it seems the function covers y from -infinity to infinity; though the mark scheme says
"x>2f(x)>2x>2 \therefore f(x)>2"

FpxGeaf4.jpg


What you gave me x2x\neq 2 is what is termed the maximal domain. That is, it's the largest set of valid x-values. Technically, when you're creating a function, any set of numbers that are valid can be used as a domain.

So summing up, the domain of a function can either be the maximal domain (which is what you are asked to find implicitly in questions that ask you to find the domain) or they can be subsets of the maximal domain which is normally given to you in a question.

In this case, the domain given is a subset of the maximal domain, namely x>2x>2. So the graph has to start as x=2 and continue to x=infinity. In which case the function starts from (2,infinity) and goes to (infinity, 2) so the range is 2 < f < infinity, or just f > 2. Infinity is implicit there. Note that the co-ordinates I gave aren't technically true. The function is undefined at 2, but the limit as x -> 2 is infinity, and the limit as x -> infinity is 2.

This is a plot of the function: http://www.wolframalpha.com/input/?i=plot+%282x-3%29%2F%28x-2%29+from+x%3D2+to+infinity
(edited 8 years ago)

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