VCE Methods Question Thread!

[cosine]

How do I solve this?



For the following find two values of x in the range 0 ≤ x ≤ 360:
sin x◦ = −0.3

Thanks !

This cannot be done by hand as 0.3 is not an exact value I.e its angle is not one of the obvious ones we are required to memorise.

[knightrider]

The braces indicate a set. Technically, a domain is a set, so rather than being equal to 2, which is a number, it is equal to {2}, which is a set that contains the number 2.

Thanks kinslayer 

[chansena]

This cannot be done by hand as 0.3 is not an exact value I.e its angle is not one of the obvious ones we are required to memorise.

Thanks for that cosine! That's why i couldn't figure it out, by hand !!

[Redoxify]

The largest possible domain of f so the g(f(x)) is defined, given f(x)=x^2 - 1 and g(x) = 3-x, x greater than or equal to 3

[cosine]

The largest possible domain of f so the g(f(x)) is defined, given f(x)=x^2 - 1 and g(x) = 3-x, x greater than or equal to 3

First of all we have to define the function f(g(x)).

For the function f(g(x)) to be defined, the range of g(x) must be a subset of the domain of f(x). The range of g(x) > 0

So for f(g(x)) to be defined, all numbers greater than zero must 'fit' inside f(x). The domain of f(x) is all real numbers, so g(x)>0 is a subset, therefore the largest domain is all Real numbers too.

It's probably confusing but try to graph the functions and it might be easier. Or you can also actually solve
f(g(x)) = (3-x)^2-1  and look at the domain from there, which is obviously all Real too.

EDIT: I did f(g(x)) instead of g(f(x)), woops

[Redoxify]

First of all we have to define the function f(g(x)).

For the function f(g(x)) to be defined, the range of g(x) must be a subset of the domain of f(x). The range of g(x) > 0

So for f(g(x)) to be defined, all numbers greater than zero must 'fit' inside f(x). The domain of f(x) is all real numbers, so g(x)>0 is a subset, therefore the largest domain is all Real numbers too.

It's probably confusing but try to graph the functions and it might be easier. Or you can also actually solve
f(g(x)) = (3-x)^2-1  and look at the domain from there, which is obviously all Real too.

I didn't understand any of that

[cosine]

I didn't understand any of that

Go over the composite functions chapter in your textbook, it should help you understand it

[Redoxify]

The largest possible domain of f so the g(f(x)) is defined, given f(x)=x^2 - 1 and g(x) = 3-x, x greater than or equal to 3

anyone show me worked solution please?

[knightrider]

How would you rotate a sin graph 45 degrees clockwise ?

[lzxnl]

How would you rotate a sin graph 45 degrees clockwise ?

You wouldn't. VCE maths doesn't ask you to
I mean if you REALLY want to, the transformation is (x', y') = (cos 45 x + sin 45 y, -sin 45 x + cos 45 y) but why would you need that?

[knightrider]

You wouldn't. VCE maths doesn't ask you to
I mean if you REALLY want to, the transformation is (x', y') = (cos 45 x + sin 45 y, -sin 45 x + cos 45 y) but why would you need that?

Thanks lzxnl 

How did you work that out?

[qwerty101]

help on this one plz , i seem to get conflicting answers

just help on part b

[keltingmeith]

Thanks lzxnl 

How did you work that out?

By mapping the variables x and y as a vector in R^2, and applying the standard rotation matrix R with angle 45 degrees. (I would like to repeat the part where this is beyond methods level)

help on this one plz , i seem to get conflicting answers

just help on part b

We want g(f(x)) to exist - this means that the range of f (that is, all the numbers that come out of f) must be a subset of the domain of g (that is, all the numbers that go into g). So, the only numbers that can go into g is in its maximal domain is x+1>=0, or x>=-1

This means, that 1/x >= -1 (that is, the range of 1/x must be a subset of the above domain) Solving this algebraically, we get x<=-1, however by inspection of the graph, you'll find that the entire right branch of the hyperbola is also >=-1 (in fact, the whole thing is strictly positive). This shows that when solving inequalities, always sketching the graphs are a good idea.

So, this means that for the range of f to be a subset of the domain of g, we require x<=-1 and x>0 (note: we do not include the zero, as 1/x is not defined for x=0 )

[lzxnl]

Thanks lzxnl 

How did you work that out?

Well...there are squillions of ways of doing this.
One way is to use polar coordinates and show that if you rotate (r, t) to (r, t+u), then your (x,y) transforms as given.
Another way is to use the linearity of rotation, which immediately gives you the entries in the transformation matrix by seeing how the rotation affects the R^2 basis vectors (1,0) and (0,1)

But again, this isn't VCE level. So don't worry.

[knightrider]

By mapping the variables x and y as a vector in R^2, and applying the standard rotation matrix R with angle 45 degrees. (I would like to repeat the part where this is beyond methods level)

Well...there are squillions of ways of doing this.
One way is to use polar coordinates and show that if you rotate (r, t) to (r, t+u), then your (x,y) transforms as given.
Another way is to use the linearity of rotation, which immediately gives you the entries in the transformation matrix by seeing how the rotation affects the R^2 basis vectors (1,0) and (0,1)

But again, this isn't VCE level. So don't worry.

Thanks guys 

What if a question came up in a sac asking us to rotate it ?