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secondary 4 | A Maths
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please help with qn 15&16. thank you
Date Posted: 3 years ago
Views: 436
For qn 16, it is the same idea but in this case since they have given you a min point.
A min point means the dy/dx = 0 at that point. From there you can obtain the equation for y. (Same method as qn 15)
as for part (ii) the idea is to use find all solution satisfying dy/dx = 0, follow by a 2nd derivative test to see which solution gives a maximum.
A min point means the dy/dx = 0 at that point. From there you can obtain the equation for y. (Same method as qn 15)
as for part (ii) the idea is to use find all solution satisfying dy/dx = 0, follow by a 2nd derivative test to see which solution gives a maximum.
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Hopefully this is clear
Date Posted:
3 years ago
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