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secondary 4 | A Maths
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Good afternoon! How do i do qns 35? Just part i will do :)
Date Posted: 4 years ago
Views: 185
Hi Candice! We show that dy/dx < 0 for all x.
Another way is to do long division and reasoning.
y = (16 - 3x) / (x + 3)
dy/dx
= (x + 3) (-3) - (16 - 3x) (1), whole thing divided by (x + 3)^2
= -9 divided by (x + 3)^2
(x + 3)^2 is always non negative but here it cannot equal zero since it is in denominator
So dy/dx < 0 for all x except 3
*correction to previous line: except -3*
Another way is to do long division and reasoning.
y = (16 - 3x) / (x + 3)
dy/dx
= (x + 3) (-3) - (16 - 3x) (1), whole thing divided by (x + 3)^2
= -9 divided by (x + 3)^2
(x + 3)^2 is always non negative but here it cannot equal zero since it is in denominator
So dy/dx < 0 for all x except 3
*correction to previous line: except -3*
I will pen this down at a later time when I am free and if no one has written them yet
It is ok! I understand it now :D Thank u for the quick response!
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Alternate method:
For a decreasing function, show dy/dx is always negative for all relevant values of x.
For a decreasing function, show dy/dx is always negative for all relevant values of x.
Date Posted:
4 years ago
Ok thank u! :)
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