VCE Methods Question Thread!

[Yacoubb]

ok you can long divide but if you're just trying to sketch it that's not necessary.



If we allow x to approach infinity, the two fractions approach 0 and become negligible. Hence, there is an asymptote at , or y=(coefficient of x in numerator)/(coefficient of x in denominator)

The other asymptote is when the function is undefined, or when the denominator equals zero. Hence, there is another asymptote at x=1/2

From this info you can just work out the intercepts and sketch the graph.

Yeah the question asks to firstly get the equation into the form I mentioned before, and then asks to find the inverse, and sketch it. So, if I wanted to sketch that, would one asymptote be at x=1/2 and the other at y=0? I'm sorry this is so foreign atm haha! I need help!! :/ thanks!

[psyxwar]

Yeah the question asks to firstly get the equation into the form I mentioned before, and then asks to find the inverse, and sketch it. So, if I wanted to sketch that, would one asymptote be at x=1/2 and the other at y=0? I'm sorry this is so foreign atm haha! I need help!! :/ thanks!

There'd be one at y=1/2, and one at x=1/2.

I'll do the working for the long division here:

, we can see that 0.5 (2x-1)s go into x+3. As 0.5(2x-1) is x-1/2, x+3-(x-1/2)=7/2, which is our remainder.

Hence, we have 1/2 r 7/2, or writing it in a more familiar form, , which is

[Yacoubb]

There'd be one at y=1/2, and one at x=1/2.

I'll do the working for the long division here:

, we can see that 0.5 (2x-1)s go into x+3. As 0.5(2x-1) is x-1/2, x+3-(x-1/2)=7/2, which is our remainder.

Hence, we have 1/2 r 7/2, or writing it in a more familiar form, , which is

Yeah that makes so much more sense. Thanks psyxwar! I really need to learn to use LaTex!!

[b^3]

If had to find the rate of change of volume with respect to height given that and that at the point where h=2, why does it make a difference if I substitute first and differentiate the expression and then substitute h=2 in, rather than substituting r for (the value of r when h is 2) and differentiating that?

As you have in terms of , the two are linked, one changes as the other changes according to the relationship you're given. If you differentiate one while you've substituted the other in as a constant, then the second isn't going to change relative to the first, as you've made it constant (think about what happens to if you just substitute in the constant, is this rate actually representing what is happening? (no)). So you need to keep everything in terms of everything else, differentiate and apply the chain rule (which is really why you can't just use the constant term first, see what happens here in the two situations), and then substitute in the constants.

[psyxwar]

(no)

wow spoilers.

Thanks!

[Snorlax]

Umm
What's the range of ??

[brightsky]

domain = [-sqrt(3),sqrt(3)]
range = [0,sqrt(3)]

[Yacoubb]

domain = [-sqrt(3),sqrt(3)]
range = [0,sqrt(3)]

Could you please explain how you got this? This function seems new to me!

[soNasty]

find all values of x for

9*e^x + 2*e^-x = 9

[Snorlax]

domain = [-sqrt(3),sqrt(3)]
range = [0,sqrt(3)]

Oh, I'll write up the question instead.

Let f : R → R, f (x) = x^2 and let g: {x: x ≤ 3} → R, g(x) =√3 − x.
State whether fog and gof exists.

I seem to be getting conflicting answers..so frustrating..

[brightsky]

The function f(x) = sqrt(a^2 - x^2) is a semicircle centred at the origin with a radius of length a.

find all values of x for

9*e^x + 2*e^-x = 9

9e^x + 2e^(-x) = 9
9e^(2x) + 2 = 9e^x (multiply both sides by e^x)
9e^(2x) - 9e^x + 2 = 0
Let A = e^x:
9A^2 - 9A + 2 = 0
the rest should be pretty straightforward

[soNasty]

The function f(x) = sqrt(a^2 - x^2) is a semicircle centred at the origin with a radius of length a.

9e^x + 2e^(-x) = 9
9e^(2x) + 2 = 9e^x (multiply both sides by e^x)
9e^(2x) - 9e^x + 2 = 0
Let A = e^x:
9A^2 - 9A + 2 = 0
the rest should be pretty straightforward

sorry i get it now, though, why did u multiply both sides by e^x? to get rid of the negative?
can you even do that? (sounds rebellious)

i got it. 2e^(-x) is 2/e^x...

[soNasty]

how would i graph something like
y=|2x+1|+|2x-1|?

these modulus graphs are confusing ..

[Nato]

how would i graph something like
y=|2x+1|+|2x-1|?

these modulus graphs are confusing ..

you could try it with addition of ordinates which is so annoying though.

[brightsky]

Let f(x) = |2x+1| and g(x) = |2x-1|. Then y = f(x) + g(x).

Sketch f(x) and g(x) on the same set of axes. Then use addition of ordinates to obtain the graph of y = f(x) + g(x).

In case you have not encountered addition of ordinates before, it is simply a method which can be used to sketch functions of the form y = f(x) + g(x). The steps involved are:
1. Sketch f(x) and g(x) on the same set of axes.
2. Choose an x-coordinate. Add the y-coordinates of f(x) and g(x) at the x-coordinate you chose, and plot the point. Then choose another x-coordinate. Add the y-coordinates of f(x) and g(x) at the second x-coordinate you chose, and plot the point. Continue until you have plotted a sufficient number of points for you to be able to make out the shape of y = f(x) + g(x).