VCE Methods Question Thread!

[knightrider]

Multiply the whole expression by -1, leaving the power to 3/5 only:



Now we know if we raise to the power of negative 1, you must reciprocate it, which will get 32/243

and then simplify further

Thanks cosine 

[knightrider]

How would you do this question?

[Special At Specialist]

How would you do this question?

It's just the intersection of the 2 domains.
If the domain of f is (-inf, 1) and the domain of g is [-1, inf), then:
Domain of (f + g) = (-inf, 1) ∩ [-1, inf)
Domain of (f + g) = [-1, 1)

Think of it like 2 conditions on a function and both conditions need to be satisfied at the same time.

[IntelxD]

You are asking me, what I asked this thread.. what are you doing with your life haha

Right now I am trying to assist people with math. No need for hurtful comments. Let's enjoy math together.

Anyhow, back to the initial dilemma. So, Mr Cosine... The two things I despise the most are rote learning and getting spoon-fed. Thus when attempting to help others I try to illuminate the path and guide them, rather than take the step for them. With this example here, I provided a very basic scenario which requires rudimentary arithmetic skills. I have no doubt that if you attempted the problem carefully you would've solved it.

So we are attempting to calculate a=0/0. Using our elite transposing skills we can multiply both sides of the equation by 0 to obtain a*0=0. Now, what number multiplied by 0 gives us 0? That's right, any number. Ta-da! We have now been introduced to a concept known as 'Indeterminate form'. This is why we can not simply take the reciprocal of both sides and assume that k=0. Unfortunately I am running out of time so if this information is insufficient I am deeply sorry. I can however direct you to websites such as wikipedia or mathsisfun to answer any remaining queries and bolster your skills in mathematics.

[knightrider]

It's just the intersection of the 2 domains.
If the domain of f is (-inf, 1) and the domain of g is [-1, inf), then:
Domain of (f + g) = (-inf, 1) ∩ [-1, inf)
Domain of (f + g) = [-1, 1)

Think of it like 2 conditions on a function and both conditions need to be satisfied at the same time.

Thanks Special At Specialist 

BTW when are we went to learn how to do these sorts of questions or these particular concepts?

[keltingmeith]

Thanks Special At Specialist 

BTW when are we went to learn how to do these sorts of questions or these particular concepts?

It should've been one of the first things you learned when you first covered functions in 3/4 methods.

[warya]

I don't know what I'm interrupting here but...

Is this a trick question?
f(x)=x^2+3 and g(x)=√(1-x)   

Establish the largest domain that will allow g(f(x)) to exist.

I'm confused though because how can we make the range of f(x)= to dom g i.e. (-infinity,1) if its [3,infinity) to start with, as in how can you restrict it to those numbers if it can't even equal those numbers anyways?

And when I tried to restrict the domain it obviously didn't work or more likely I'm doing something wrong?

Any help is appreciated!

[warya]

Yeah but for that statement to be made doesn't the domain of f(x) have to be restricted because the range needs to be a subset of the domain of g which is (-infinity,-1)? Because once its established to exist, only then does the domain of the composite function g(f(x)) becomes the domain of the inside function

Idk maybe I'm just confusing myself

I just can't see how (3,infinity) is made to be a subset of (-infinity,-1)

Edit: grad year 2018 Cosine? Joke orrr

[IntelxD]

Yeah but for that statement to be made doesn't the domain of f(x) have to be restricted because the range needs to be a subset of the domain of g which is (-infinity,-1)? Because once its established to exist, only then does the domain of the composite function g(f(x)) becomes the domain of the inside function

Idk maybe I'm just confusing myself

I just can't see how (3,infinity) is made to be a subset of (-infinity,-1)

Edit: grad year 2018 Cosine? Joke orrr

The values for the range of f and domain of g you have obtained are both correct (both 3 and -1 should be inclusive though). Some functions just can not be composed so unless you have made a mistake with the actual equations, it is a trick question. Thinking of it conceptually, any output value (range) you obtain from f(x) will be equal to or greater than 3. Subbing these values into g(x) will never give you a real solution. In other words, there is no interval for the range of f(x) that is a valid input value for g(x) (within the domain of g(x)). You can think of functions like f(x)=-x^2 and g(x)=log(x). The composition g(f(x)) will never exist.

[warya]

The values for the range of f and domain of g you have obtained are both correct (both 3 and -1 should be inclusive though). Some functions just can not be composed so unless you have made a mistake with the actual equations, it is a trick question. Thinking of it conceptually, any output value (range) you obtain from f(x) will be equal to or greater than 3. Subbing these values into g(x) will never give you a real solution. In other words, there is no interval for the range of f(x) that is a valid input value for g(x) (within the domain of g(x)). You can think of functions like f(x)=-x^2 and g(x)=log(x). The composition g(f(x)) will never exist.

Thanks man that makes sense

[cosine]

Thanks man that makes sense

Why does the domain of two functions added together become the intersection of the two?

[keltingmeith]

Why does the domain of two functions added together become the intersection of the two?

h(x) = f(x) + g(x)

f(x)=2 for 1<=x<=3
g(x)=1 for 2<=x<=4

So:

h(2)=f(2)+g(2)=2+1=3
h(3)=f(3)+g(3)=2+1=3
h(4)=f(4)+g(4)=?+1=?
h(1)=f(1)+g(1)=2+?=?

Basically, the intersection of the domains is the only time both are defined, and hence the only time their sum/difference can be defined.

[cosine]

h(x) = f(x) + g(x)

f(x)=2 for 1<=x<=3
g(x)=1 for 2<=x<=4

So:

h(2)=f(2)+g(2)=2+1=3
h(3)=f(3)+g(3)=2+1=3
h(4)=f(4)+g(4)=?+1=?
h(1)=f(1)+g(1)=2+?=?

Basically, the intersection of the domains is the only time both are defined, and hence the only time their sum/difference can be defined.

I sorta get it now, so what if there is no intersection of the two functions, would there be a domain??

[keltingmeith]

I sorta get it now, so what if there is no intersection of the two functions, would there be a domain??

Nope. Consider the function - try sketching it on your CAS/google, see what comes out.

[lzxnl]

Nope. Consider the function - try sketching it on your CAS/google, see what comes out.

Technically you shouldn't see anything as you can't 'see' a point in 2D