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March 29, 2024, 10:39:51 am

Author Topic: VCE Methods Question Thread!  (Read 4802909 times)  Share 

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guac

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Re: VCE Methods Question Thread!
« Reply #17670 on: February 14, 2019, 07:52:38 pm »
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This is modulus and is not in the methods course.

In any case,
log e |x^2 - 5x - 6 | = 0 actually implies that |x^2 - 5x - 6| = 1,  since log e (1) = 0
thus you will want to solve both x^2 - 5x - 6 = 1 and x^2 - 5x - 6 = -1 because of the absolute sign

The modulus part may not be relevant to the methods course, but please take note that log e (a) = 0 does not imply that a = 0
« Last Edit: February 14, 2019, 07:55:23 pm by guac »
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persistent_insomniac

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Re: VCE Methods Question Thread!
« Reply #17671 on: February 16, 2019, 11:26:49 am »
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Are matrices actually in the methods 3/4 course or do we just need to understand basic multiplication, solving equations to apply them to transformations?

addict

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Re: VCE Methods Question Thread!
« Reply #17672 on: February 16, 2019, 12:03:33 pm »
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Are matrices actually in the methods 3/4 course or do we just need to understand basic multiplication, solving equations to apply them to transformations?

Nope. You don't even need to understand "basic multiplication, solving equations to apply them to transformations", though they're very good to know.
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EllingtonFeint

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Re: VCE Methods Question Thread!
« Reply #17673 on: February 16, 2019, 12:58:31 pm »
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Hey,
I'm finding jt hard to find the inverse of
g:[-1,infinity) --> R, g(x) =x ^2 +2x

I keep ending up with (+-sqrtx/2 )/2 which is NOT right :o

Could somebody please explain what I'm doing wrong and how to solve this problem please?
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17674 on: February 16, 2019, 01:19:24 pm »
+3
Hey,
I'm finding jt hard to find the inverse of
g:[-1,infinity) --> R, g(x) =x ^2 +2x

I keep ending up with (+-sqrtx/2 )/2 which is NOT right :o

Could somebody please explain what I'm doing wrong and how to solve this problem please?
\[g:[-1,\infty)\to\mathbb{R},\ g(x)=(x+1)^2-1\]Let \(y=g^{-1}(x)\).
\begin{align*}x&=(y+1)^2-1\\
x+1&=(y+1)^2\end{align*} Now, \(\text{dom}(g)=\text{ran}(g^{-1})=[-1,\infty)\), so \[y+1=\sqrt{x+1}\ \text{only.}\] \[\text{Thus, }\ g^{-1}:[-1,\infty)\to\mathbb{R},\ g^{-1}(x)=\sqrt{x+1}-1.\]
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17675 on: February 16, 2019, 03:19:26 pm »
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Does \(\tan\left(-7\right)\) = \(\tan^{-1}\left(7\right)\) and if so, why? Is there a law to this? Will this work to both sin and cos where the negative sign of an answer transfer to a \(\sin^{-1}\) and \(\cos^{-1}\) to a positive answer? Thanks

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17676 on: February 16, 2019, 04:15:17 pm »
+2
Does \(\tan\left(-7\right)\) = \(\tan^{-1}\left(7\right)\) and if so, why? Is there a law to this? Will this work to both sin and cos where the negative sign of an answer transfer to a \(\sin^{-1}\) and \(\cos^{-1}\) to a positive answer? Thanks

There's clearly some confusion about notation here. The function \(\tan^{-1}(.)\) (which in VCE is more commonly notated as \(\arctan(.)\)) denotes the inverse function of \(\tan(.)\). In general, \(\tan(a)\neq\arctan(a)\).

I'm not sure what you mean by your second question. For example, \(-\arcsin(-1)=\dfrac{\pi}{2}>0\) (if this is what you're asking).

However, you should note that inverse circular functions are not in the Methods course.

Edit: are you trying to refer to odd and even functions perhaps?
« Last Edit: February 16, 2019, 09:10:09 pm by dantraicos »
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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17677 on: February 16, 2019, 06:29:06 pm »
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There's clearly some confusion about notation here. The function \(\tan^{-1}(.)\) (which in VCE is more commonly notated as \(\arctan(.)\)) denotes the inverse function of \(\tan(.)\). In general, \(\tan(a)\neq\arctan(a)\).

I'm not sure what you mean by your second question. For example, \(-\arcsin(-1)=\dfrac{\pi}{2}>0\) (if this is what you're asking).

However, you should note that inverse circular functions and not in the Methods course.

Edit: are you trying to refer to odd and even functions perhaps?
Umm, I don't think you get what I mean.
I'm just wondering if \(\tan\left(-7\right)=\tan^{-1}\left(7\right)\) and whether this works with both cos and sin aswell. >.<

RuiAce

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Re: VCE Methods Question Thread!
« Reply #17678 on: February 16, 2019, 06:34:53 pm »
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Umm, I don't think you get what I mean.
I'm just wondering if \(\tan\left(-7\right)=\tan^{-1}\left(7\right)\) and whether this works with both cos and sin aswell. >.<
You can check on the calculator to see that they are not equal.

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17679 on: February 16, 2019, 07:39:34 pm »
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You can check on the calculator to see that they are not equal.
If I do tan(-7), it becomes -tan(7), and now because this isn't \(\tan^{-1}\), how do I make it become that?

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17680 on: February 16, 2019, 08:00:44 pm »
+2
If I do tan(-7), it becomes -tan(7), and now because this isn't \(\tan^{-1}\), how do I make it become that?

I'm not actually sure what you mean by "making it become that".

I think you might be having trouble using the CAS calculator. When you type any expression, your CAS tries to keep it in exact form. To overcome that, instead of pressing enter, press ctrl+enter.

In each of the images below, the first calculation is obtained by pressing enter only. The second one is with ctrl+enter.



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aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17681 on: February 16, 2019, 08:18:37 pm »
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ok
well cna someone help me put
\(\tan\left(-7\right)\) in the terms of \(\tan^{-1}\) ? I'm only wanting to know if  \(\tan\left(-7\right)=\tan^{-1}\left(7\right)\)

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Re: VCE Methods Question Thread!
« Reply #17682 on: February 16, 2019, 08:36:49 pm »
+4
ok
well cna someone help me put
\(\tan\left(-7\right)\) in the terms of \(\tan^{-1}\) ? I'm only wanting to know if  \(\tan\left(-7\right)=\tan^{-1}\left(7\right)\)

We already told you that they're not equal?

Also, you cannot express \( \tan (-7)\) in terms of \(\tan^{-1}\). Note that \( \tan (-7) = -\tan 7\) as you said, and that's it. We cannot bring inverse tan into this.

aspiringantelope

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Re: VCE Methods Question Thread!
« Reply #17683 on: February 16, 2019, 09:29:46 pm »
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We already told you that they're not equal?

Also, you cannot express \( \tan (-7)\) in terms of \(\tan^{-1}\). Note that \( \tan (-7) = -\tan 7\) as you said, and that's it. We cannot bring inverse tan into this.
thanks for ur answer
and no one told me it wasnt equal before

AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17684 on: February 16, 2019, 09:33:12 pm »
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thanks for ur answer
and no one told me it wasnt equal before

We did, and I sent CAS screenshots showing they weren't equal:
In general, \(\tan(a)\neq\arctan(a)\).

But that doesn't matter. I hope everything's been cleared up now :)
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