How do I find the domain and range of these questions? (1 Viewer)

[DrDawn]

The ones you posted? Find if any values are excluded from the domain. This includes denominator being 0, and square roots being negative. Then use calculus to find turning points and get a rough idea of the graph to find the range. These are mainly principles, I don't know if this would be considered a 'clear' method. I always preferred to use graphs if it would make sense and wouldn't take too long

Yeah it's pretty straight forward finding the domain and range from a graph but without a graph it's pretty difficult to find the range

 

[DrDawn]

You're best looking at a textbook for this.

Which textbook would you recommend?

 

[DrDawn]

Can someone help find the domain and range for question a), I'm a bit confused what effect the numerator has as it is "x" instead of "1"

 

[DrDawn]

Tbh I don't think the numerator being x matters.

I checked the graph and your domain is correct, but the range is wrong

 

[DrDawn]

I'm a bit confused what effect the numerator has as it is "x" instead of "1"

@Trebla could you please explain what effect x has as a numerator on the range (the domain would still be the same even if there was a "1" as the numerator but the range changes) thanks!

 

[Trebla]

@Trebla could you please explain what effect x has as a numerator on the range (the domain would still be the same even if there was a "1" as the numerator but the range changes) thanks!

It depends on the function. In the particular example you are referring to, it expands the range, namely it allows negative and zero y-values to be possible.

If you compare



which has a range of all real y

and



which has a range of y > 0

then the multiplication of the y-values (i.e. any number times a positive number) expands the range to all real y.

This is something you learn in Extension 1 though. In Advanced, it is expected you use calculus to figure out the graph.

 

[DrDawn]

Let me try some more for good measure, hopefully I haven't fallen asleep yet.
b) f(x) = 2/(x^2-4)
let y = f(x)
Domain: x can't be -2 nor 2
I think the range is all real y excluding y = 0 because I think I know what this graph should (roughly) look like.
Oh no I drew the graph roughly and I'm wrong! The range is y > 0, y <= -1/2 (g0d I put -1/4 at first because I'm asleep now)

I think I woke up after that one lol

c) 1/(x^2+x)
Domain: x can't be 0 nor -1
Range: x > 0, x <= -4

d) 1/((x-3)(x-2))
Domain: x can't be 3 nor 2
Range: x > 0, x <= -4

e) sqrt(x^2-4)
Domain: (x-2)(x+2) >= 0 so that is x > 2, x < -2
Range: y >= 0

f) 1/(sqrt(1-x^2))
Domain: 1-x^2 > 0 --> -1 < x < 1
Range: *gets calculator to plug in values* y >= 1

I checked them with desmos so I hope they're right now

Thanks! I understand how you got the domain but I'm still a little confused how you got the range, could you please explain that?