VCE Methods Question Thread!

[deStudent]

http://m.imgur.com/a/C3pvs

For photo 4, part f) how do we find the that the domain of the inverse is (2,infinite)? My working is attached.

Photo 2, part c) for their second line. They used polynomial long division to find the numerator it seems, but why do they still divide by 3x-2? Their numerator is in the form Q(x)P(x) + R(x), do they divide this by 3x-2 so it still equals the original expression?

Thanks

[lzxnl]

http://m.imgur.com/a/C3pvs

For photo 4, part f) how do we find the that the domain of the inverse is (2,infinite)? My working is attached.

Photo 2, part c) for their second line. They used polynomial long division to find the numerator it seems, but why do they still divide by 3x-2? Their numerator is in the form Q(x)P(x) + R(x), do they divide this by 3x-2 so it still equals the original expression?

Thanks

4f
Well...let's find the range. By definition, fourth roots are never negative. So, the first term must be positive.

2c
Basically, they want to divide the numerator such that they can separate anything with an x in it. They want to have a constant as the numerator.

[Shadowxo]

http://m.imgur.com/a/C3pvs

For photo 4, part f) how do we find the that the domain of the inverse is (2,infinite)? My working is attached.

Photo 2, part c) for their second line. They used polynomial long division to find the numerator it seems, but why do they still divide by 3x-2? Their numerator is in the form Q(x)P(x) + R(x), do they divide this by 3x-2 so it still equals the original expression?

Thanks

4 f) - in the original function, the range is (2,infinity) as if the denominator is large (if x is large) then y approaches 2, and as x approaches 1 (denominator approaches zero) then y approaches infinity. the range of the initial function = domain of inverse function. You can only find the inverse function where it's 1:1, the function you found isn't 1:1 therefore you can't find the domain using it, as it's restricted by the range of the original function.

2. c) Since both the x's are to the power of 1 and there aren't any x's, they're trying to get it into the form a + b/(cx+d). They're doing this so there's only one x, so they can easily find the inverse function. They took 2/3 out of the x so it would become 3x, then they subtracted a 2 and added a 2, which expanded out to +4/3, which is where that came from. They basically put the function into a different form with a single x to make finding the inverse function easier.

Does this make sense? If you need further explanation just let me know

[deStudent]

^Shadowxo, thanks a lot!

For this question http://m.imgur.com/a/f50vV
Part a) I don't get how proving ran f = dom f shows that fof is defined? Wouldn't it be defined for R but x =1/2 regardless of the range? Confused on why the range is required to show its defined essentially.

b) to find the restriction of x, do we have to consider the line where no simplifications are made, ie the first line of working?

c) for the easier method, how does fof(x) = x, make sense? Isn't that implying that f is the inverse of f(x), but aren't they the same function?

[Shadowxo]

^Shadowxo, thanks a lot!

For this question http://m.imgur.com/a/f50vV
Part a) I don't get how proving ran f = dom f shows that fof is defined? Wouldn't it be defined for R but x =1/2 regardless of the range? Confused on why the range is required to show its defined essentially.

b) to find the restriction of x, do we have to consider the line where no simplifications are made, ie the first line of working?

c) for the easier method, how does fof(x) = x, make sense? Isn't that implying that f is the inverse of f(x), but aren't they the same function?

a) fog is defined if the range of g is a subset of f (in this case they're both f). Just because it's f o f doesn't mean it'll be defined: eg -√x, f o f won't be defined unless the domain is restricted to 0. So finding the range of f lets you know if f o f is defined. In this case, the range is R\{1/2}. The domain is R\{1/2} so it's defined, as ran(f) is a subset (in this case equal to) dom(f)
b) yes, because although you can simplify it, it still won't exist if anything is divided by zero. An example: if you were given \(\frac{x^2}{x}\) you could simplify it to x but x couldn't be equal to zero.
c) I believe this only works in specific scenarios. You've found f o f (x) which equals x (it often won't though). f(f^-1x) always = x, and this time f(f(x)) = x, so f^-1(x) = f (x). In this case, this function is its own inverse.

Hope this helps a bit

[problempolly]

Wondering if someone could help me with this! I'm a bit stuck!


Question:

"The number of bacteria can be modelled by y(t) = 2*5^(t)

Show how to express this equation in the form y(t) = e^(ct+d)"



Thanks heaps 

[Shadowxo]

Wondering if someone could help me with this! I'm a bit stuck!


Question:

"The number of bacteria can be modelled by y(t) = 2*5^(t)

Show how to express this equation in the form y(t) = e^(ct+d)"



Thanks heaps 

2 = e^d
5 = e^c
Sub in
y(t) = e^d*(e^c)^t
y(t) = e^ct+d

If they want you to find c and d
d = log_e2 (they may want you to approximate)
c = log_e5

Alternatively:

Hope this helps

[problempolly]

Thanks Shadowxo! 

[Shadowxo]

Thanks Shadowxo! 

Happy to help! 

[deStudent]

a) fog is defined if the range of g is a subset of f (in this case they're both f). Just because it's f o f doesn't mean it'll be defined: eg -√x, f o f won't be defined unless the domain is restricted to 0. So finding the range of f lets you know if f o f is defined. In this case, the range is R\{1/2}. The domain is R\{1/2} so it's defined, as ran(f) is a subset (in this case equal to) dom(f)
b) yes, because although you can simplify it, it still won't exist if anything is divided by zero. An example: if you were given \(\frac{x^2}{x}\) you could simplify it to x but x couldn't be equal to zero.
c) I believe this only works in specific scenarios. You've found f o f (x) which equals x (it often won't though). f(f^-1x) always = x, and this time f(f(x)) = x, so f^-1(x) = f (x). In this case, this function is its own inverse.

Hope this helps a bit

Thanks.

For c) is there anyway to tell f is its own inverse by observation, so I can avoid all the working that I had?

[Shadowxo]

Thanks.

For c) is there anyway to tell f is its own inverse by observation, so I can avoid all the working that I had?

Normally they wouldn't ask you to do that.
The inverse of a function is when it's flipped across the y=x line. If a function is symmetrical along the line y=x, then the function will be its own inverse
In this case, you'd need to rearrange it first like in part a, to f(x) = \(\frac{1}{2}+ \frac{7}{2(2x-1)}\). From this you know its general shape is 1/x, and the asymptotes are x=1/2 and y=1/2. From this you can tell its symmetrical along the line y=x and is therefore its own inverse.
Don't worry about this though, you'd probably only have to tell if a function is its own inverse if you were to do all the working you've had to do (and even then I haven't encountered it)

[deStudent]

Having some problems with these questions http://m.imgur.com/a/6tKZv

Q3b) what's the best way to find the domain? I was initially pretty lost but would considering x - 12>0 and y-20>0 be best? Also im not sure why the answer has the domain as exclusive 12 but inclusive 60. Shouldn't they both be exclusive?

4b) why don't we need to state the domain of V? The answer only gave the domain of x.

Q6d) how do we determine the shape? Area = 6a/(a+2).

Thanks

[Shadowxo]

Having some problems with these questions http://m.imgur.com/a/6tKZv

Q3b) what's the best way to find the domain? I was initially pretty lost but would considering x - 12>0 and y-20>0 be best? Also im not sure why the answer has the domain as exclusive 12 but inclusive 60. Shouldn't they both be exclusive?

4b) why don't we need to state the domain of V? The answer only gave the domain of x.

Q6d) how do we determine the shape? Area = 6a/(a+2).

Thanks

Hi,
3b) Yes, y ≥ 20 and x ≥ 12. The perimeter is 160m², so x + y = 80, y = 80-x. therefore 80 - x ≥ 20, x ≤60. And I think they should both be inclusive as x=12, y=68 would work, and so would x=60, y=20. The question's kind of ambiguous, they should both be the same though. If they specify that it cannot be a rectangle / x must be greater than 12 and y must be greater than 20 then they should both be exclusive. If they don't specify, it should be inclusive.
4b)They state "The volume is fixed" so it's a constant, therefore you don't need to do a domain for it.
6d) rearrange it so it doesn't have a on the top.

Then sketch

Hope this helps

[Perryman]

hey,
i need help on this question...
For the simultaneous equations mx + 2y = 2 and 8x + my = m, find the values of m for which there are infinitely many solutions.

so far i hav solved the equations to remove y, and then i substituted the values into the discriminant formula: delta=b^2 - 4ac
but i am not sure whether that is right or where to go from there..
please help!

thanks!!

[keltingmeith]

hey,
i need help on this question...
For the simultaneous equations mx + 2y = 2 and 8x + my = m, find the values of m for which there are infinitely many solutions.

so far i hav solved the equations to remove y, and then i substituted the values into the discriminant formula: delta=b^2 - 4ac
but i am not sure whether that is right or where to go from there..
please help!

thanks!!

I'm not going to say what you're doing is wrong - however, I'm not sure if it would arrive at the right answer. At the same time, it's not how I'd approach the question.

I'd note that if there are infinitely many solutions, then the lines must touch each other at every point - that is, they're the same line.

So, I'm going to solve both for y:



Now that they're in this form, it's obvious that they have the same y-intercept, 1. So, this means that the two lines will be equal when the gradients are equal:



Therefore, this set of simultaneous equations will have infinitely many solutions when m=-4,4

In hindsight, eliminating y was the correct step, but I think that using the discriminant would not have helped you at all. It depends how you went about removing y, and what quadratic you used.