VCE Methods Question Thread!

[Yueni]

How do I solve the attached question?

Derive and sub in the point in question.

[Quantum44]

8e^-x-e^x=2 solve for x

Multiply both sides by e^x then let a = e^x and now it is a simple quadratic to solve

[zhen]

Does anyone if for strictly increasing/decreasing intervals are we allowed to include the x value of turning points given that it is either at the start/end of the interval?

Yea, I'm pretty sure stationary points are included in strictly increasing and decreasing intervals. But, you should check with someone else just in case I'm wrong.

[Sine]

Does anyone if for strictly increasing/decreasing intervals are we allowed to include the x value of turning points given that it is either at the start/end of the interval?

Yes in terms of VCE maths methods strictly increasing and decreasing include points where the gradient is 0 e.g. a Turning point

[clarke54321]

Can someone please help me out with this question?

If g(x)=log2(x-2) and 2[g(x)]=g(f(x)) then f(x) is equal to:    (sorry about the log equation, it should be base 2)

TIA 

[Sine]

Can someone please help me out with this question?

If g(x)=log2(x-2) and 2[g(x)]=g(f(x)) then f(x) is equal to:    (sorry about the log equation, it should be base 2)

TIA 

g(x)=log₂(x-2)

LHS =2[g(x)]
        = 2log₂(x-2)
        = log₂((x-2)²)

RHS = g(f(x))
         = log₂((f(x)-2)

let's now combine the equation

LHS = RHS
log₂((x-2)²)  =  log₂((f(x)-2)
The interior of the logs must be equal.
(x-2)² = (f(x)-2
Hence
f(x) = (x-2)² +2
 

[zhen]

Can someone please help me out with this question?

If g(x)=log2(x-2) and 2[g(x)]=g(f(x)) then f(x) is equal to:    (sorry about the log equation, it should be base 2)

TIA 

g(f(x))=2log2(x-2)
g(f(x))=log2(x-2)^2
g(f(x))=log2(x^2-4x+4)
g(f(x))=log2(x^2-4x+6-2)
Therefore f(x)=x^2-4x+6
This is what I got, but it might be wrong, so someone else should check.

Edit: Sine beat me to it

[Guideme]

the line perpendicular to the graph y=g(f(x)) where f(x)=1/x and g(x)=-1/(x^2)+x is given by y=-x+a where a is a real constant. Find the possible values of a

[Syndicate]

the line perpendicular to the graph y=g(f(x)) where f(x)=1/x and g(x)=-1/(x^2)+x is given by y=-x+a where a is a real constant. Find the possible values of a

1. find g(f(x) (= -x^2 + 1/x)
2. find the non-perpendicular gradient (which is 1)
3. differentiate g(f(x)), and solve for x, when dy/dx = 1 (x=-1)
4. Calculate the y-coordinate of g(f(x)) (y = -2)
5. sub the coordinates into the equation of the perpendicular line ( y = -x + a)
6. you should get a = -3

[simrat99]

How do transformations of a function affect its domain? And does anyone know how to change the x values on the table of a graph on the TI-nspire calculator? Thanks 

[Azzzz]

Hello all!
There is a question that has a few students in my year confused.
The question is "If Luke's weight increases linearly from birth at the same rate as that between 6 and 24 weeks, what is Luke's predicted weight at 40 weeks?
The original birth weight is 3.9 kg. At 6 weeks the weight is 5.5 kg and at 24 weeks it's 8 kg. Two of my friends whom are both at the top of the cohort for Methods both did it in a linear form (y=mx+c) and got like 10.2 kg by putting the gradient between 6 and 24 weeks where m is and 8 where y is and 24 where x Is and then solved for C. Once they solved for C they multiplied the average rate between 6-24 weeks by 40 and added whatever they got for C onto that. However, all I did was multiply the average rate between 6-24 weeks by 40 and added the original birth weight and got around 9.45 kg. Sorry if any of this confused you guys but how would you solve this question?
Thanks

[Yueni]

Hello all!
There is a question that has a few students in my year confused.
The question is "If Luke's weight increases linearly from birth at the same rate as that between 6 and 24 weeks, what is Luke's predicted weight at 40 weeks?
The original birth weight is 3.9 kg. At 6 weeks the weight is 5.5 kg and at 24 weeks it's 8 kg. Two of my friends whom are both at the top of the cohort for Methods both did it in a linear form (y=mx+c) and got like 10.2 kg by putting the gradient between 6 and 24 weeks where m is and 8 where y is and 24 where x Is and then solved for C. Once they solved for C they multiplied the average rate between 6-24 weeks by 40 and added whatever they got for C onto that. However, all I did was multiply the average rate between 6-24 weeks by 40 and added the original birth weight and got around 9.45 kg. Sorry if any of this confused you guys but how would you solve this question?
Thanks

Going by the question itself and how it's worded, the equation should be in the form of y=mx+3.9, where m is the gradient between 6 and 24 weeks.

[Yueni]

How do transformations of a function affect its domain? And does anyone know how to change the x values on the table of a graph on the TI-nspire calculator? Thanks 

Dilations from y affect the domain the same way it does the function. Translations along x move the domain.

[Syndicate]

How do transformations of a function affect its domain? And does anyone know how to change the x values on the table of a graph on the TI-nspire calculator? Thanks 

The domain is only affected by the translation of x units along the x-axis, or being dilated across the y-axis. I hope someone can help with your your TI-nspire question (I am not the best at using it either).

Hello all!
There is a question that has a few students in my year confused.
The question is "If Luke's weight increases linearly from birth at the same rate as that between 6 and 24 weeks, what is Luke's predicted weight at 40 weeks?
The original birth weight is 3.9 kg. At 6 weeks the weight is 5.5 kg and at 24 weeks it's 8 kg. Two of my friends whom are both at the top of the cohort for Methods both did it in a linear form (y=mx+c) and got like 10.2 kg by putting the gradient between 6 and 24 weeks where m is and 8 where y is and 24 where x Is and then solved for C. Once they solved for C they multiplied the average rate between 6-24 weeks by 40 and added whatever they got for C onto that. However, all I did was multiply the average rate between 6-24 weeks by 40 and added the original birth weight and got around 9.45 kg. Sorry if any of this confused you guys but how would you solve this question?
Thanks

Firstly, write down all the information you know.
t (days) = 0, w (kilograms) = 3.9
From this you should already be able to tell that c = 3.9
m = the rate of change in his weight and age
m = (8-5.5)/126
= 5/252

So therefore w = 5/252 t + 3.9
at t = 40 x 7
w = 5(40 x 7)/252 +3.9
w = 9.46 kg

I used days as if I set my y variable in weeks, the points would be inconsistent with the line. So therefore you are correct. The method you used is essentially what your friends did, however, they disregarded the initial weight information which would have led them to getting the wrong answer.

EDIT: I'll post my solution anyways as I was in the process of doing it when Yueni posted his 

[LPadlan]

number 5