# BoS Maths Trials 2019 (1 Viewer)

##### -insert title here-
Math isn't my area but isn't this the case???????
trivial counterexample: x+y=0 and the inverse x+y=0 are tangent at (0, 0) with slope -1.

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#### Drdusk

##### Moderator
Moderator
Either way here's how I would do it

$\bg_white y = a^x \Rightarrow \dfrac{dy}{dx} = ln(a)a^x$

$\bg_white \therefore ln(a)a^p = m \hspace{2mm}\text{but}\hspace{2mm} a^p = p \Rightarrow pln(a) = m\hspace{2mm} (1)$

$\bg_white y = log_a(x)\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{xln(a)}\Rightarrow m = \dfrac{1}{pln(a)}\hspace{2mm} (2)$

$\bg_white \text{Subbing (1) into (2)}$

$\bg_white \therefore \dfrac{1}{m} = m\Rightarrow m^2 = 1$

##### -insert title here-
The intended solution:

Inverse Function Theorem:

Let b = f(a), f differentiable around a

(f⁻¹)'(b) = 1/f'(a)

Regarding the problem:

By the Inverse Function Theorem, (f⁻¹)'(p) = 1/f'(p)

But the slopes are equal, by the problem assumption.

So let m = f'(p) = (f⁻¹)'(p), and we have m = 1/m ⇒ m² = 1

##### -insert title here-
Either way here's how I would do it

$\bg_white y = a^x \Rightarrow \dfrac{dy}{dx} = ln(a)a^x$

$\bg_white \therefore ln(a)a^p = m \hspace{2mm}\text{but}\hspace{2mm} a^p = p \Rightarrow pln(a) = m\hspace{2mm} (1)$

$\bg_white y = log_a(x)\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{xln(a)}\Rightarrow m = \dfrac{1}{pln(a)}\hspace{2mm} (2)$

$\bg_white \text{Subbing (1) into (2)}$

$\bg_white \therefore \dfrac{1}{m} = m\Rightarrow m^2 = 1$
While this solution is correct, it is much more cumbersome to work with, especially since you are pulling it down to a specific case. If it was put in a more general form, then it would be slightly less cumbersome to write out.

But then at that point it would just be easier to use the Inverse Function Theorem to kill it in one hit.

#### integral95

##### Well-Known Member
While this solution is correct, it is much more cumbersome to work with, especially since you are pulling it down to a specific case. If it was put in a more general form, then it would be slightly less cumbersome to write out.

But then at that point it would just be easier to use the Inverse Function Theorem to kill it in one hit.

I honestly don't recall learning the inverse function theorem until first year uni to be fair (and basically never used it from 2nd year and later) , so I doubt the majority of the high-school students would even consider it.

Please share some more rants pls this is hilarious.

##### -insert title here-
apparently 5 people think that

$\bg_white \frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1}$

##### -insert title here-
I honestly don't recall learning the inverse function theorem until first year uni to be fair (and basically never used it from 2nd year and later) , so I doubt the majority of the high-school students would even consider it.

Please share some more rants pls this is hilarious.
you don't remember learning slope of inverse function at inverse point = 1/(slope of function at function point) ?

because i certainly do remember that class in 3U

#### TheOnePheeph

##### Active Member
you don't remember learning slope of inverse function at inverse point = 1/(slope of function at function point) ?

because i certainly do remember that class in 3U
I'm pretty sure its not in the syllabus. There have been hsc questions on it before, but I definitely haven't learned it in class either

##### -insert title here-
I'm pretty sure its not in the syllabus. There have been hsc questions on it before, but I definitely haven't learned it in class either
well fortunately, the other solution also works out (and a fair few students did go about that way solving it)

##### -insert title here-
Either way here's how I would do it

$\bg_white y = a^x \Rightarrow \dfrac{dy}{dx} = ln(a)a^x$

$\bg_white \therefore ln(a)a^p = m \hspace{2mm}\text{but}\hspace{2mm} a^p = p \Rightarrow pln(a) = m\hspace{2mm} (1)$

$\bg_white y = log_a(x)\Rightarrow \dfrac{dy}{dx} = \dfrac{1}{xln(a)}\Rightarrow m = \dfrac{1}{pln(a)}\hspace{2mm} (2)$

$\bg_white \text{Subbing (1) into (2)}$

$\bg_white \therefore \dfrac{1}{m} = m\Rightarrow m^2 = 1$
actually i think this is not the only way to do it using differentiation, but it is probably the most obvious now that i think about it from a student perspective

#### Trebla

I'm pretty sure its not in the syllabus. There have been hsc questions on it before, but I definitely haven't learned it in class either
It is in the syllabus, but in a slightly different format. Students are expected to be taught that
(dy/dx)(dx/dy) = 1
so they can then do reciprocals of derivatives (e.g to solve differential equations). This is basically how you find the derivative of the inverse function.

#### TheOnePheeph

##### Active Member
It is in the syllabus, but in a slightly different format. Students are expected to know that
(dy/dx)(dx/dy) = 1
so they can do reciprocals of derivatives. This is basically how you find the derivative of the inverse function.
Well damn, we skipped completely over that in class haha. Still not too difficult to derive luckily, and is a fairly basic concept of inverse functions.

#### Trebla

Well damn, we skipped completely over that in class haha. Still not too difficult to derive luckily, and is a fairly basic concept of inverse functions.
You were probably taught it implicitly without knowing. It basically formalises the ability to take reciprocals of derivatives - for example when finding the solution to dN/dt = kN

#### TheOnePheeph

##### Active Member
You were probably taught it implicitly without knowing. It basically formalises the ability to take reciprocals of derivatives - for example when finding the solution to dN/dt = kN
Yeah well it is pretty much an application of that, which is how they are able to ask questions on it in the hsc. Its more we weren't specifically told:

$\bg_white \frac{d}{dx} f^{-1}\left(x\right) = \frac{1}{f^{'}\left(f^{-1}\left(x\right)\right)}$

Oh well, its not hard to derive anyway lol.

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#### StudyOnly

##### Active Member
I was taught this special result because of MIF

#### Drdusk

##### Moderator
Moderator
I was taught this special result because of MIF
Bruuuh, and that people is the pinnacle of wtf?

##### -insert title here-
Yeah well it is pretty much an application of that, which is how they are able to ask questions on it in the hsc. Its more we weren't specifically told:

$\bg_white \frac{d}{dx} f^{-1}\left(x\right) = \frac{1}{f^{'}\left(f^{-1}\left(x\right)\right)}$

Oh well, its not hard to derive anyway lol.
except based on what Trebla said, that wasn't even need anyway, I'm just an idiot who hasn't bothered reviewing the syllabus outlines

by the statement of the question, one can see that dy/dx (p,p) = dx/dy (p,p) = 1

and the conclusion follows

#### TheOnePheeph

##### Active Member
except based on what Trebla said, that wasn't even need anyway, I'm just an idiot who hasn't bothered reviewing the syllabus outlines

by the statement of the question, one can see that dy/dx (p,p) = dx/dy (p,p) = 1

and the conclusion follows
Ah right, I didn't see that he meant that. Yeah thats much easier than the other method haha.

#### Drdusk

##### Moderator
Moderator
apparently 5 people think that

$\bg_white \frac{\text{d}}{\text{d}n} \left( \left(1+\frac{1}{n}\right)^n \right) = \left(-\frac{1}{n^2}\right) n \left( 1+\frac{1}{n} \right)^{n-1}$
Haha what really?