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The f'(x) is going down so it is increasing a decreasing rate

The graph displayed is f'(x), the gradient function. Assuming that it has a horizontal asymptote at y = 0, then it never touches the x-axis. That means f'(x) > 0, so always positive! Doesnt matter if the f' graph is swirling/jumping crazily as long as it is > 0 for all real x. Positive gradient...

(i) Increasing. The sign of the derivative shows when the curve is increasing (>0) or decreasing (<0) because derivative is the gradient function. i.e. input a point and it outputs the gradient at the point to the curve or gradient of tangent. Increasing: gradient always positive so curve is...

Re: HSC 2015 4U Marathon - Advanced Level Are you posting your homework?

i dont watch it but why say that?

His integral is 1+x instead of 1+x^2 for denominator

My bad, the picture you posted looked like it had a diameter (chord to complete triangle) lol.

Yep, if you look at your diagram and label the arc lengths, you can see that your diagram demonstrates 5y < 4y

I think your diagram is incorrect. The chord completing the triangle should be to the right of the centre because of the arc length ratios

It can't go through the centre. Arc length on each side has to be equal for that construction to cross the centre

Since v is a real number, then v^2 is always positive and is at least 0. C = 1000 + 10v^2 \geq 1000 + 2(0) = 1000 That is the same as C \geq 1000 Which means that C is at least 1000.

Is your theta to the left or right of the centre? Is it facing 5y or x?

\sin ^2 x + \cos^2 x = 1 Divide both sides by sin^2 x, 1+ \cot ^2 x = \cosec^2 x

Re: HSC 2015 3U Marathon Maximise LHS by maximising RHS and how do you maximise RHS? (hint: fraction) Ding dong, calculus problem without needing to go nuts on differentiating that junk mindlessly.

Re: UNSW Chit Chat Thread 2015. ahh yeah, that's an easier way so yep you are correct on saying Real < 0 and Im = 0 for the points where log is not analytic

Re: UNSW Chit Chat Thread 2015. why? taking the log of a negative yields an undefined value so not analytic

Re: UNSW Chit Chat Thread 2015. yeah there are more if you consider the log curve in real and extend to complex, here are the areas where log is analytic Re(z + 1/z + 2) > 0 and Im(z + 1/z + 2) != 0

Then why add a nugatory comment at the beginning? You say it because you want to but do you need to? EDIT: I was being a dick, apologies.

Why don't you know why? Surely the teacher would be happy to tell you where your errors lie.. Maybe you need to revise HSC verbs? Like evaluate means adv+disadv+conclusion on which side you want to take

Re: UNSW Chit Chat Thread 2015. Where is Log(z + z^{-1} + 2) analytic?