Maths Help! (1 Viewer)

[Kujah]

We're doing functions and log in Maths atm, and is there any quick way to find the range of a function without drawing it or creating a table of values.

eg:

y = x / x-1

 

[Riviet]

Check for horizontal asymptotes in the function as these will nearly all the time affect your range.

Dividing top and bottom by highest power of x (x¹ in this case), you should observe that the horizontal asymptote is y=1.

So the range should be all real y, y=/=1.

Another useful way is by making x as a function in terms of y:

Given y=x/(x-1),

y(x-1)=x

xy-y=x

xy-x=y

x(y-1)=y

x=y/(y-1)

Clearly, y=/=1 but y is defined for all other real values so this is your range.

Note that this takes some time and does not always work.

 

[simonloo]

Yeah, what he said.

 

[SoulSearcher]

You'll catch up.

 

[Kujah]

Another useful way is by making x as a function in terms of y:

Given y=x/(x-1),

y(x-1)=x

xy-y=x

xy-x=y

x(y-1)=y

x=y/(y-1)

Clearly, y=/=1 but y is defined for all other real values so this is your range.

Note that this takes some time and does not always work.

I find this method much more easier to understand.

 

[Kujah]

So for:

y= 1 / x^2 - 3x + 2

D: Any real x-value, except for x= 2 or 1
R: By making x as a function in terms of y, I got the range can be any real value except for y=0.

Is that right?

 

[toadstooltown]

For this one it's going to be harder to make x the subject. Limits aren't all that hard, just a bit weird to get used to at first.
y=1/(x²-3x+2)
y=1/[(x-2)(x-1)]. As the denominator can not =0, x=/=1,2, but true for all other x.
So domain is x=/=1,2, but all other belonging to the reals.

For range we divide the whole fraction by the highest power of x in denominator, as Rivet said (x²)
y=[1/x²]/[x²/x² -3x/x²+2/x²]
y=[1/x²]/[1-3/x+2/x²]
We know the special limit results that Lim (x->0) of 1/x is infinity and that Lim (x->inf) of 1/x is 0.
So we do the lim as x approches infinity.
y=[0]/[1-0+0]
So y approaches 0, never reaching it.
So range, y=/=0, all other reals.

Eventually you'll be able to just look at the function and know the domain and range within a second or two, so don't worry!

 

[davidbarnes]

You're already doing Logs??? We've only just finished learning Pythagoras/Trig.

 

[PrettyVacant]

You're already doing Logs??? We've only just finished learning Pythagoras/Trig.

Weren't you supposed to do those in Yr 9 or something?

 

[SoulSearcher]

He's in a different state

 

[davidbarnes]

No idea, although we only learnt them this year in the oast few weeks. My school is not the best, lol.