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topics / holidays completion (1 Viewer)
- Thread starter hasterz
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sydneyphoenix
Member
Math during holiday
I've more or less gone through all the 3U stuff, and would spend rest of the holiday doing past papers.
As for 4U, I think I would pass over mechanics, volume and harder 3U for a time being, and go through Cambridge on other chapters.
By the way, which text do you reckon's better, Fitzpatrick or Cambridge?
I've more or less gone through all the 3U stuff, and would spend rest of the holiday doing past papers.
As for 4U, I think I would pass over mechanics, volume and harder 3U for a time being, and go through Cambridge on other chapters.
By the way, which text do you reckon's better, Fitzpatrick or Cambridge?
How can you draw this?
1. f(x) = cosx.e^-x
Can you please draw this in paint.
2. f(t) = 5/ ( 2+3.e^-t) , t equals to or more than 0.
Show that f ' (t) > 0 for all values of t in the domain. How could you do this?
Do we show 0 doesn't work but works for 1, 2, 3....?
1. f(x) = cosx.e^-x
Can you please draw this in paint.
2. f(t) = 5/ ( 2+3.e^-t) , t equals to or more than 0.
Show that f ' (t) > 0 for all values of t in the domain. How could you do this?
Do we show 0 doesn't work but works for 1, 2, 3....?
Slidey
But pieces of what?
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For question 1, look up graphing by multiplication of ordinates in the Graphs topic.Tuna said:
For question 2...
f(t)=5.(3.e^-t + 2)^-1, t>=0
(3.e^-t+2)'=-3.e^-t
.'. f'(t)=5*-1*-3.e^-t*(3.e^-t+2)^-2
f'(t)=15.e^-t/(3.e^-t+2)^2
Numerator is positive. Denominator is positive. Numerator can never equal zero, .'.' f'(t) > 0
grimreaper
Member
haha by the end of these holidays last year I hadnt even completed a single 4u topic.. we were still doing complex numbers. And we were behind in 3u as well
Trev
stix
we hav to hav polynomial chapter finished - urgh 
