VCE Methods Questions Thread

Hey guys,

How do you find the domain of:
√x-1/x+2 ?

We know that x-1/ x+2 has to be >= 0

I tried doing it in a way that I brought x+2 to the other side and then I had:
x-1 >= 0

But the answer says:
x <-2 U x >=1

How...
Is there a method to do this?

Please help.

Thanks.

PizzaMaster

Function:
√(x-1)/( x+2)

√(x-1) implies that x≥1 as you cannot have a negative value inside the root (to get a real value). x+2 is the denominator and cannot equal zero. So the domain of x+2 would be any real number except -2. But since the domain of √(x-1) is x≥1, we can disregard the -2 and be left with the final answer of x≥1. I'm not sure if I interpreted your function correctly tho, my answer is correct tho according to my interpretation. Let me know if the function is listed above is the one you are talking about, I CAS verified it as well.

Thank you for your help.
However, I meant that the (x-1) and (x+2) are both under the square root sign.
Sorry for the confusion.

No worries, in that case we have a whole new situation. The same principle applies in that x+2 cannot equal zero and everything inside the bracket must be 0 or greater. To be honest, I would convert (x-1)/(x+2) into a partial fraction which would yield 1 - 3/(x+2) which can be obtained through long division. If you graph that you get a hyperbola (I think it's called that). You just gotta find when the graph is greater than or equal to zero. I would strongly suggest graphing it and you'll come to realise that from -2<x<1 the graph is negative. Thus the domain of that function would be the direct opposite, aka your aforementioned answer.

You'll find graphing stuff is always gonna help you and you need to learn and get used to do doing it. Hope that Helps.

PizzaMaster Nice to meet a fellow pizza lover

PizzaMaster
If you have f(x) = x⁰.5

Then f-1(x) = x², but restricted to the domain R+ u {0}, which means the inverse is also 1-1. The negative branch of x² isn't a reflection of y = x⁰.5 in the line y = x, so it's not part of the inverse function

Ohh so you mean that the range of the original is the domain of the inverse?

And so that's why
f(x) = 2x^ 3/4 + 1

does have an inverse:
f-1(x) = [(x-1) / 2]^ 4/3
BUT with domain [1, infinity)

How do we know whether the question wants us to give the dilation from x axis or y axis when given both the original and transformed function.

For example:
f(x)= 1/x²
f1(x)= 5/x²

I got dilation by a factor of 1/5 from the y axis but the answers say factor of 5 from the x axis. Is my answer still valid?

Billzene

Ohh ok but how do we whether to work out it’s dilation from the X or Y axis if it doesn’t say in the question?

snowflake by recognition of the form y = a/(1/b(x-h))ⁿ + k

a = dilation factor from x axis
b = dilation factor from y axis
h = translation in +ve direction of x axis
n = power
k = translation in the positive direction of y axis

This is similar to turning point form of parabolas

Billzene

thank youuu

find values of a and b such that a(x+2)+b(x+3)=18x+8 for all values of x
if someone could help that would be great

chemistry1111 Expand then factorising LHS gives (a+b)x + (2a + 3b) = 18x + 8 --> by comparing coefficients, you obtain a+b = 18 and 2a+3b =8, solve simultaneously for a and b, giving a=46 and b=-28

thank you so much that makes sense

also this question
For the polynomial P(x)=(a+1)x³ + (b-7)x² +c+5, find values for a,b and c if P(x) has a degree 2, a leading coefficient of 3 and the constant term is -1

when factorizing polynomials in methods unit 3/4, what method do most people use vcaa prefer, long division or equating coefficients

Do we need to buy the new Methods 3/4 book for the new study design (2023)?
I feel like the textbook questions are the same...