VCE Methods Question Thread!

[Perryman]

Hi Perryman,
The domain would be R. (found in part b.)

But for that to occur you cant really sketch the graph in such a small area as it has a lot of ups and downs, and there isnt really a pattern, correct me if i am wrong!!.........maybe i hav the wrong equation but when i sketch -cos(4e^x) it doesnt look that wonderful to try and plot!!!

[scout]

But for that to occur you cant really sketch the graph in such a small area as it has a lot of ups and downs, and there isnt really a pattern, correct me if i am wrong!!.........maybe i hav the wrong equation but when i sketch -cos(4e^x) it doesnt look that wonderful to try and plot!!!

I got the same equation. You're right, it's an awful graph! I've just double-checked with my CAS; it also says that the domain of f(g(x)) is R. How strange... I'm sorry, I'm afraid I can't help in this case  . ...Unless there are any extra conditions given; it's very unusual that there isn't a specific domain... for a graph without a clear pattern as you rightly say!

I wonder, where did you get this question?

[Perryman]

I got the same equation. You're right, it's an awful graph! I've just double-checked with my CAS; it also says that the domain of f(g(x)) is R. How strange... I'm sorry, I'm afraid I can't help in this case  . ...Unless there are any extra conditions given; it's very unusual that there isn't a specific domain... for a graph without a clear pattern as you rightly say!

I wonder, where did you get this question?

it was jst on a booklet that we got from out teacher.....i think he may hav made it up himself.......
but yeh thts all the information that we get sooo......i wasnt sure!!......also for some reason my calculator reckons that it is undefined after about x=29......but this didnt really make sense.......so i just put R down as the domain...cause that is the answer to most trig functions.....!....but yeh dont know why it doesnt continue....i think the values stuffed the calculator up.....because if you sub in 50 for x it refuses to give you a proper answer......bit strange...might just ask my teacher next class........unless someone can help furthermore!!!

[lzxnl]

it was jst on a booklet that we got from out teacher.....i think he may hav made it up himself.......
but yeh thts all the information that we get sooo......i wasnt sure!!......also for some reason my calculator reckons that it is undefined after about x=29......but this didnt really make sense.......so i just put R down as the domain...cause that is the answer to most trig functions.....!....but yeh dont know why it doesnt continue....i think the values stuffed the calculator up.....because if you sub in 50 for x it refuses to give you a proper answer......bit strange...might just ask my teacher next class........unless someone can help furthermore!!!

Let's think what happens when you sub in x = 50.

cos(4e^50). Hmm. 2^50 < e^50 < 3^50. 2 is roughly 10^0.3 and 3 is roughly 10^0.5 (very, very roughly)
So e^50 is between 10^15 and 10^25. Yikes.

One really important problem you have for this, practically, is because the calculator will have an error in the value of e (it has to use a finite number of decimal places), its value for e^x will also have an error, and for larger x values, this error becomes so significant that your value for cos(4e^x) isn't reliable anymore.

[TheCommando]

Hey guys, im having trouble with 15b, 16a and 17b
Really dont understand it at all
http://imgur.com/a/nfxzF
Ps does any know why i cant upload one image because apparently its way to big which it isnt

[zhen]

Hey guys, im having trouble with 15b, 16a and 17b
Really dont understand it at all
http://imgur.com/a/nfxzF
Ps does any know why i cant upload one image because apparently its way to big which it isnt

This is for 16a. Right now I feel really tired, so I just did the fastest one. I'll probably answer the rest later if no one answers it by the next time I'm on.

[keltingmeith]

Hey guys, im having trouble with 15b, 16a and 17b
Really dont understand it at all
http://imgur.com/a/nfxzF
Ps does any know why i cant upload one image because apparently its way to big which it isnt

16a, see above. However, for the other two, it simply comes down to turning English into maths. Think about what dy/dx corresponds to, and what each X value refers to.

I could just give you the answer, but it's not going to help your own analytical thinking or problem solving skills. If you give these questions a shot (hell, have a stab in the dark), then I can help you with any misunderstandings you might have.

[MisterNeo]

Hey guys, im having trouble with 15b, 16a and 17b
Really dont understand it at all
http://imgur.com/a/nfxzF
Ps does any know why i cant upload one image because apparently its way to big which it isnt

Sorry for the late reply...
For 17B, it asks you to find the mass at t=0, then find at which t-value does that mass equal half of the t=0 value.

Let t=0


Find t when m=1

[ringring]

Hey can anyone help me with this question

The graph of y = ax^2 + bx has a turning point at (1,-2). The values of a and b are:

[Syndicate]

Hey can anyone help me with this question

The graph of y = ax^2 + bx has a turning point at (1,-2). The values of a and b are:

eqn 1)
1. differentiate the function.
dy/dx = 2ax + b

2. Set it equal to 0, and substitute in x = 1.
0 = 2a + b
eqn 2)
1. Substitute in x = 1, and y = -2
-2 = a + b
Now solve simultaneously.
-2 = a + b (1)
0 = 2a+b (2)
(2)-(1)
a = 2
b = -4

[ringring]

Thanks!
What about this one, I'm a bit lost

The gradient graph of the cubic function with rule y = f(x) crosses the x-axis at (1, 0)
and (-2, 0). The maximum value of the gradient function is 6. The value of x for which
the graph of y = f(x) has a local minimum is:

[Syndicate]

Thanks!
What about this one, I'm a bit lost

The gradient graph of the cubic function with rule y = f(x) crosses the x-axis at (1, 0)
and (-2, 0). The maximum value of the gradient function is 6. The value of x for which
the graph of y = f(x) has a local minimum is:

If the gradient graph (as in the derivative) crosses the x-intercept, that means there is a local maximum or a local minimum on the actual graph. So when f'(x) = 0, there is a local max/min on f(x).

I am assuming no other information is given (I am also going to assume that this is a positive cubic function). Therefore the local min is at x = 1, as the max occurs more to the left of the x-axis.

Look below for the correct explanation

[TheCommando]

Hey guys, how do i do Q6
https://postimg.org/image/z8ipy2f9f/

[Guideme]

Q16 and 17 pls

[zhen]

Hey guys, how do i do Q6
https://postimg.org/image/z8ipy2f9f/

If f(x) is the function and it touches 2y+6x=15 at (0,15/2) and has a stationary point at (3,-6). Then it means that f(0)=15/2, f(3)=-6, f'(3)=0 and f'(0)=gradient of the line. From that you get simultaneous equations and solve for a,b,c and d. Hope that helps.