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It'd still be possible to have a function's graph have infinite slope somewhere yet be continuous there, e.g. The graph of the cube root function at the origin. A place where the slope is infinite makes the function non-differentiable there (even if it is continuous there) because the derivative...

Re: HSC 2016 2U Marathon $\noindent Let $f(x)=mx+b$, where $m,b$ are real constants.$ $\noindent (i) Show that $f(x+1)-f(x)= f^\prime (x)$.$ $\noindent (ii) Does the above result surprise you? If not, why not?$

In general, what we need (and is sufficient) to conclude that 0 is the only solution is that the columns of the matrix are what is called linearly independent (in the case of a square matrix, we can also replace the word 'columns' in the above with 'rows'). Basically what this means is that all...

No. E.g. If the system was: x + y = 0 2x + 2y = 0 (So the matrix is: [1 1] [2 2] ) there are clearly infinitely many solutions.

Yes. Alternatively, you could just give it as R – {3}. Like how if the domain was all reals, you could just write the domain as R.

Some of the variants are pretty fun (my favourite would probably be Crazyhouse), but I think they deteriorate your regular chess ability if you don't keep practising that as well.

That thread is now locked.

Has anyone here ever played Crazyhouse chess (or any other variants) on lichess? Lol.

$\noindent One way is to parameterise the line as: $x=t,y=2-\frac{2}{3}t$, for $t\in \mathbb{R}$. This comes from rewriting the line equation as $y=2-\frac{2}{3}x$. Now, the squared-distance of a point with $x=t$ on the line from the origin is: $S(t) = t^2 + \left(2-\frac{2}{3}t\right)^2$. So...

Imagine if it was promotion of all pawns to knights instead. :haha:

Some questions do get asked very often, e.g. things like asking to derive the equation of the tangent to an ellipse at an arbitrary given point.

You deduct everything. I realised davidgoes4wce didn't write his deduction of super, but the final answer he got includes this deduction (so he probably just forgot to type the super amount, i.e. typo).

That is correct. The answers are here: https://www.boardofstudies.nsw.edu.au/hsc_exams/2015/guides/2015-hsc-mg-maths-general-2.pdf .

Re: MATH1131 help thread Idk, by saying that the limits are 0 from the left and right, it seems (in my opinion) too close to assuming what they're asking you to prove.

Re: MATH1131 help thread I did it (proved the required limit) via the epsilon-delta definition of a limit.

Re: MATH1131 help thread Note f(0) = 0, so we need to show that f(x) -> f(0) = 0 as x -> 0. So we need to show that given eps > 0, there exists a delta > 0 such that if 0 < |x-0| < delta (i.e. 0 < |x| < delta), then |f(x) - 0| < eps. To do this, let delta = eps, for any given eps > 0. Then...

How many years back are we allowed to go?

Re: MATH1131 help thread To sketch y ≤ 2x, first sketch y = 2x. This is the boundary of the region. Then the places where y is less than (or equal to) 2x will be the places below (or on) this line. If the restriction is 0 ≤ x ≤ 2, then we only take the part where x is between 0 and 2. To...

Well essentially the motivation behind making delta to be less than 2 is to deal with the |x+1| in the |x+1||x-1| stage. When we're at |x+1||x-1|, we want to make this less than eps. We can control |x-1| directly, since it's less than delta, so we can make it less than eps/4 say (we could also...

Delta is the smaller of the two. The importance of this is that if we make something less than delta, it'll be simultaneously less than 2 and eps/4.