Original post by officialzak_2001
Help on last question
If is increasing then , and it's decreasing then
RDKGames explained it well. Use that fact to solve the problem.
Original post by officialzak_2001
Help on last question
If is increasing then
Original post by officialzak_2001
Help on last question
If f(x) is increasing over [-1,1] that means for -1<=x<=1 f'(x) >= 0.
So 2x+p >= 0 for those values of x.
Plugging in -1 --> p >= 2.
(edited 6 years ago)
Original post by thekidwhogames
If f(x) is increasing over [-1,1] that means for -1<=x<=1 f(x) >= 0. f'(x) ?
So 2x+p >= 0 for those values of x.
Plugging in -1 --> p >= 2. ?? -1 is not correct
So 2x+p >= 0 for those values of x.
Plugging in -1 --> p >= 2. ?? -1 is not correct
, the inequality is strict.
Original post by BobbJo
, the inequality is strict.
It’s only strict if the function is strictly increasing.
Though at AS level you dont know this difference so OP may as well use the strict version as that’s probably what the question intends
Original post by RDKGames
It’s only strict if the function is strictly increasing.
Though at AS level you dont know this difference so OP may as well use the strict version as that’s probably what the question intends
Though at AS level you dont know this difference so OP may as well use the strict version as that’s probably what the question intends
Oh I see
Original post by thekidwhogames
If f(x) is increasing over [-1,1] that means for -1<=x<=1 f'(x) >= 0.
So 2x p >= 0 for those values of x.
Plugging in -1 --> p >= 2.
So 2x p >= 0 for those values of x.
Plugging in -1 --> p >= 2.
Why can’t you sub in 1 instead of -1 and get p is greater or equal to -2
(edited 6 years ago)
Original post by officialzak_2001
Why can’t you sub in 1 instead of -1 and get p is greater or equal to -2
You can, and that gives as you say. But if you consider the fact that at we must have as well then we obtain a stricter condition on which is
Clearly, the range of values which satisfy both inequalities and can be represented by
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