VCE Methods Question Thread!

[spectroscopy]

noob as question but can someone please show me how to solve for x using LCD not cross multiplication
(7x+3)/2 = (9x-8)/4

[Nato]

noob as question but can someone please show me how to solve for x using LCD not cross multiplication
(7x+3)/2 = (9x-8)/4

well, you could multiply the numerator and denominator of the left hand side by 2. this will result in the denominator of the fraction on the LHS being 4, equal to the one of the left hand side.

you could then multiply both sides by four, and be left with 2(7x+3)=9x-8 to solve

EDIT: changed to 'multiply both sides by 4, instead of divide.

[spectroscopy]

well, you could multiply the numerator and denominator of the left hand side by 2. this will result in the denominator of the fraction on the LHS being 4, equal to the one of the left hand side.

you could then divide both sides by four, and be left with 2(7x+3)=9x-8 to solve

yep that works thanks, im a bit fuzzy on my maths rules LOL cheers

[Cort]

Would anyone explain what a and b would be in |a-b|? Is it always two numbers, or could one represent the letter x (which could be any number) and the other the an actual number?
 Also - what is the relevance of the note (in the picture) in relation to drawing up the modulus? That is, why |a-b| would be switched around in the second function. Is it because there's a possibility that a that's x could be smaller than b?

Cheers,
Cort.

[b^3]

and can be numbers or variables, but in methods you'll most likely get one being a number and the other being a variable.

The relevance of the note is that you're really turning the modulus function into a hybrid function, by splitting the modulus up into two sections of the original function which it effects differently (due to one being below the axis and the other being above the axis, including zero). The modulus function will take any outputs that are negative, so below the axis (in this case the 'inside of the modulus' is negative when ), and it will 'flip' what you have in the axis so that you end up with the same magnitude of what you had, but with a positive sign (e.g. if you had -2, then you would get if you have , where you know that is negative, then you're going to flip it across the axis for the domain for which this is satisfied, that is for , ). When the original function is above the axis (or zero), that is when is positive or zero, (i.e. , then the modulus has no effect on the function and you get out exactly what you put in. That is for , so , .

Hope that helps

EDIT: I should note here that after the flip in the axis you need to remember to translate the graph 1 unit upwards.

EDIT2: Rephrased jibberish a bit.

[Nato]

some hints on solving for x:

[Stevensmay]

some hints on solving for x:

Continue isolating the ln(x) from there until you can change it to something else easily.

Solution.

Spoiler

Observe that this will be true if x = 1.

[lzxnl]

Continue isolating the ln(x) from there until you can change it to something else easily.

Solution.

Spoiler

Observe that this will be true if x = 1.

Can I just say that the question sucks?
If you ever get something like this with a log term and a separate x, if it's non-calc, the answer has to be really simple. In this case just guess, as Stevensmay suggested. You wouldn't be able to solve 2ln x + 2 = x, for instance.

It's a dud question.

[psyxwar]

Is there anyway of sketching x^3+5x besides addition of ordinates/ plotting out points etc?

[lzxnl]

Is there anyway of sketching x^3+5x besides addition of ordinates/ plotting out points etc?

y=x^3+5x
dy/dx = 3x^2 + 5
Evidently there are no stationary points
dy/dx>0 for all x => draw a graph that is always increasing.

y=0 => x=0 or x^2+5=0 => x=0 only.

dy/dx = 3x^2+5. Upon consideration you can see that the minimum gradient occurs at x=0, so make the graph not as steep at the origin. You should be able to do the rest.


PS. I never use addition of ordinates.

[GeniDoi]

Wasn't sure how to do this one, I know its multiple chain rules nested in eachother though and got up to:

Let f(x) = u
Let (x^2 + 1)^1/2 = v
dy/dx = 2*u'*u
= 2*(1+v)'(1+v)
= 2*(1+v)'(dv/dx(x^2+1)(1+v)

Not sure where to go further, do I continue chaining/am I doing it correctly?

[brightsky]

dy/dx = 2 (x + sqrt(x^2 + 1)) * (1 + (2x)/(2sqrt(x^2 + 1)))

Simplify and then sub in y.

[psyxwar]

(Image removed from quote.)

Wasn't sure how to do this one, I know its multiple chain rules nested in eachother though and got up to:

Let f(x) = u
Let (x^2 + 1)^1/2 = v
dy/dx = 2*u'*u
= 2*(1+v)'(1+v)
= 2*(1+v)'(dv/dx(x^2+1)(1+v)

Not sure where to go further, do I continue chaining/am I doing it correctly?

let y=u^2
dy/dx=2u*du/dx
du/dx=d/dx

• + d/dx[(x^2+1)^1/2]*2x

du/dx=1+1/2(x^2+1)^(-1/2)*2x
dy/dx=2u*du/dx

Expand (cbs typing all this up soz) and make it all over sqrt(x^2+1).

Your numerator is 4x^2+4x(sqrt(x^2+1))+2, which is the expanded form of 2y^2.

[b^3]

working

EDIT: Beaten e.t.c.

[Nato]

Can I just say that the question sucks?
If you ever get something like this with a log term and a separate x, if it's non-calc, the answer has to be really simple. In this case just guess, as Stevensmay suggested. You wouldn't be able to solve 2ln x + 2 = x, for instance.

It's a dud question.

haha, the question wasn't really like that. I had to graph some exponential function and the inverse. I knew they intersected at y=x, so i just made it equal to x, to find the point haha