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February 23, 2019, 10:05:23 pm

Author Topic: VCE Methods Question Thread!  (Read 2165461 times)  Share 

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Sine

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Re: VCE Methods Question Thread!
« Reply #17700 on: February 12, 2019, 05:37:24 pm »
+2


Just make sure that when you are differentiating logs the structures is if g(x) = log(f(x)) then g'(x) = f'(x)/f(x)

EllingtonFeint

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Re: VCE Methods Question Thread!
« Reply #17701 on: February 13, 2019, 07:54:53 pm »
0
Iíve just started learning inverse functions (Iím not behind, am I?!) and I donít really understand how to find the domain and range of an inverse so could somebody please explain it to me?
So for example...

For the following function find the inverse and state itís domain and range
f:[-2,0] ó> R, f(x) = 2x - 4
So, I found that the inverse is...
f^-1 (x) = 1/2 (x+4)

Iím just not sure how to find the domain and range.
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EllingtonFeint

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Re: VCE Methods Question Thread!
« Reply #17702 on: February 13, 2019, 07:55:40 pm »
0
Iíve just started learning inverse functions (Iím not behind, am I?!) and I donít really understand how to find the domain and range of an inverse so could somebody please explain it to me?
So for example...

For the following function find the inverse and state its domain and range
f:[-2,0] ó> R, f(x) = 2x - 4
So, I found that the inverse is...
f^-1 (x) = 1/2 (x+4)

Iím just not sure how to find the domain and range.
2018: Biology [49] | Indonesian SL: [40]

🌱Offering biology tutoring in Mornington Peninsula area or online 🌱

S200

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Re: VCE Methods Question Thread!
« Reply #17703 on: February 13, 2019, 07:59:38 pm »
+1
Iíve just started learning inverse functions (Iím not behind, am I?!) and I donít really understand how to find the domain and range of an inverse so could somebody please explain it to me?
So for example...

For the following function find the inverse and state itís domain and range
f:[-2,0] ó> R, f(x) = 2x - 4
So, I found that the inverse is...
f^-1 (x) = 1/2 (x+4)

Iím just not sure how to find the domain and range.
Something I found helpful to understanding this topic was to graph out the given function, the inverse function, and the function \(f(x)=x\)

Doing this, you will see that the domain of the given function is the range of the inverse function and vice versa.
Spoiler
In your equation, the Range would be \([-2,0]\) and the domain would be \([1,2]\)
« Last Edit: February 13, 2019, 08:04:51 pm by S200 »
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EllingtonFeint

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Re: VCE Methods Question Thread!
« Reply #17704 on: February 13, 2019, 08:04:41 pm »
0
Something I found helpful to understanding this topic was to graph out the given function, the inverse function, and the function \(f(x)=x\)

Doing this, you will see that the domain of the given function is the range of the inverse function and vice versa.

Iím sorry, I still donít understand...  :-[

So graph out  2x-4 , 1/2(x+4) and then what would the last one be??
I kinda get the idea of that line where it reflects but still not completely sure. Could you just explain a lil more please? :)
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S200

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Re: VCE Methods Question Thread!
« Reply #17705 on: February 13, 2019, 08:07:02 pm »
0
Iím sorry, I still donít understand...  :-[

So graph out  2x-4 , 1/2(x+4) and then what would the last one be??
I kinda get the idea of that line where it reflects but still not completely sure. Could you just explain a lil more please? :)
\(f(x)=x\) is the function of the line \(y=x\). As you guessed, this is the line of reflection for the graphs and is the reason that the \(\text {domain}^1 = \text {range} ^{-1}\)
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Re: VCE Methods Question Thread!
« Reply #17706 on: February 14, 2019, 07:45:41 pm »
0
How to solve?

log e |x^2 - 5x - 6 | = 0

|x^2 - 5x - 6 | = 0


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Re: VCE Methods Question Thread!
« Reply #17707 on: February 14, 2019, 07:50:51 pm »
0
How to solve?

log e |x^2 - 5x - 6 | = 0

|x^2 - 5x - 6 | = 0

This is modulus and is not in the methods course.

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Re: VCE Methods Question Thread!
« Reply #17708 on: February 14, 2019, 07:52:38 pm »
0
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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Re: VCE Methods Question Thread!
« Reply #17709 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?

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Re: VCE Methods Question Thread!
« Reply #17710 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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Re: VCE Methods Question Thread!
« Reply #17711 on: February 16, 2019, 12:58:31 pm »
0
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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Re: VCE Methods Question Thread!
« Reply #17712 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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Re: VCE Methods Question Thread!
« Reply #17713 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
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AlphaZero

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Re: VCE Methods Question Thread!
« Reply #17714 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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