Just a reminder this is a marathon thread, not for posting textbook errata. I have removed those posts and put into a separate thread
davidgoes4wce Useful Maths Questions & Textbook Errata 2017
Just a reminder this is a marathon thread, not for posting textbook errata. I have removed those posts and put into a separate thread
davidgoes4wce Useful Maths Questions & Textbook Errata 2017
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OK I got nothing else to do on a Friday night I am going to have an honest go at this question.
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what happened to this thread o
and how do you guys know if you thoroughly know a topic in math?
HSC 2017- 90.85
Goal: get 90s in all subjects... rip
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Im going to Year 10 in 2018 and am learning differential calculus, to answer OP's question I sketched the graph, any advanced criticism is appreciated as I am learning
When you do working in 2u make sure to write out obvious statements, they may seem stupid but they are necessary to get the marks, particularly for curve sketching. For example above- include in your working a phrase like "testing nature," and a concluding statement "therefore horizontal point of inflexion at P(X,Y) etc.
It doesn't matter if your graph's scale isn't perfect, you don't need to measure out centimetres on your ruler. As long as you sketch a reasonable graph, showing any necessary features (including any that were asked for) e.g. intercepts, maxima/minima, pts. of inflexion you'll get the marks. The intent is not to test you on your measuring skills.
Just as a quick example, I would sketch it something like this. Just note that this graph shows the y-intercept more clearly. It's possible to find any x intercepts but this isn't expected in the 2U course, although it should be in the 3U course. I was taught to label my points of inflexion with a line going through the point but I don't think it matters as long as you make it clear that the point IS one of inflexion. In an exam I would draw the graph larger, though.
A point of inflexion means a change in concavity, i.e. from positive to negative or vice versa. so to prove that a point is one of inflexion, you need to show that the sign of the second derivative changes around that point.
for your curve:
Because changes signs around , a point of inflexion exists at .
Just to elaborate on why you need to test the second derivative around the point - first note that any quadratic, such as , has no points of inflexion because is always a constant (in this case, ) and so the equation has no solution.
Consider the curve . The curvature of the graph is identical to (you can graph it digitally to check). Here, , so the equation has a solution at . But from the previous paragraph, clearly this curve can't have a point of inflexion. This is because for all real , so the second derivative never changes signs and because of this, no points of inflexion exist on this curve.
Last edited by fan96; 13 Dec 2017 at 10:53 PM.
HSC 2018 - [English Adv.] • [Maths Ext. 1] • [Maths Ext. 2] • [Chemistry] • [Software Design and Development]
_{1}∫^{(3√3)} t^{2} dt cos(3π/9) = log(^{3}√e) | Integral t^{2} dt, From 1 to the cube root of 3. Times the cosine, of three pi over nine, Equals log of the cube root of e.
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