# how to show a trigonometric function is increasing?

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#1
Hi,

here (again) yes my maths teachers want to kill me before COVID does :P

I’ll attach the question and my working - i just don’t know how to explain it? I’ve drawn a diagram to prove it and worked out dy/dx.

Thanks!
Last edited by dxnixl; 1 week ago
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1 week ago
#2
(Original post by dxnixl)
Hi,

here (again) yes my maths teachers want to kill me before COVID does :P

I’ll attach the question and my working - i just don’t know how to explain it? I’ve drawn a diagram to prove it and worked out dy/dx.

Thanks!
A function is increasing if its derivative is . So as long as you can reason out why the derivative never negative, then you're good.
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#3
(Original post by RDKGames)
A function is increasing if its derivative is . So as long as you can reason out why the derivative never negative, then you're good.
so do i just differentiate xcosx again ergh i know that but for some reason i’ve gone blank :/ What do i substitute into xcosx to prove its greater or equal to 0?
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1 week ago
#4
(Original post by dxnixl)
so do i just differentiate xcosx again ergh i know that but for some reason i’ve gone blank :/ What do i substitute into xcosx to prove its greater or equal to 0?
Not again ... you already got the derivative, you just need to explain why it is .

Note what the values of x are. Are they all positive? So is cos(x) always positive? Are you always ending up with a product xcos(x) which is positive?
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#5
(Original post by RDKGames)
Not again ... you already got the derivative, you just need to explain why it is .

Note what the values of x are. Are they all positive? So is cos(x) always positive? Are you always ending up with a product xcos(x) which is positive?
yep - if i substitute any value between 0 and Pi/2, i always get a result > 0, so it’s an increasing function.

Would writing this be enough for the marks or do i have to numerically prove it?
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1 week ago
#6
strictly speaking the gradient must be positive for an increasing function. if is zero then we say it is "stationary"
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1 week ago
#7
(Original post by dxnixl)
yep - if i substitute any value between 0 and Pi/2, i always get a result > 0, so it’s an increasing function.

Would writing this be enough for the marks or do i have to numerically prove it?
To numerically prove it you would need to substitute in every real number between 0 and pi/2 ... up for the challenge?

All you need to do is reason it out, and as long as you do that you will get the marks.
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1 week ago
#8
(Original post by the bear)
strictly speaking the gradient must be positive for an increasing function. if is zero then we say it is "stationary"
https://www.math24.net/increasing-decreasing-functions/

It's like sequences.

{ 0 , 0 , 0 , ...} is increasing because

{ 0 ,1 , 2 , ... } is strictly increasing because
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#9
(Original post by RDKGames)
To numerically prove it you would need to substitute in every real number between 0 and pi/2 ... up for the challenge?

All you need to do is reason it out, and as long as you do that you will get the marks.
Alright thanks
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1 week ago
#10
(Original post by dxnixl)
Hi,

here (again) yes my maths teachers want to kill me before COVID does :P

I’ll attach the question and my working - i just don’t know how to explain it? I’ve drawn a diagram to prove it and worked out dy/dx.

Thanks!
please double check your derivative, did you remember to use the product rule for xsinx
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1 week ago
#11
(Original post by dxnixl)
Hi,

here (again) yes my maths teachers want to kill me before COVID does :P

I’ll attach the question and my working - i just don’t know how to explain it? I’ve drawn a diagram to prove it and worked out dy/dx.

Thanks!
please double check your derivative, did you remember to use the product rule for xsinx
0
1 week ago
#12
(Original post by RDKGames)
https://www.math24.net/increasing-decreasing-functions/

It's like sequences.

{ 0 , 0 , 0 , ...} is increasing because

{ 0 ,1 , 2 , ... } is strictly increasing because
http://www.sunshinemaths.com/topics/...easing-curves/
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1 week ago
#13
(Original post by Melvin Guna)
please double check your derivative, did you remember to use the product rule for xsinx
The function is
xsin(x) + cos(x)
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1 week ago
#14
(Original post by the bear)
strictly speaking the gradient must be positive for an increasing function. if is zero then we say it is "stationary"
You can have isolated points where the gradient is zero and still have a strictly increasing function. E.g. x^3 is strictly increasing (everywhere) but has derivative 0 at x = 0. (Most A-level references won't deal with this edge case, however).
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1 week ago
#15
(Original post by mqb2766)
The function is
xsin(x) + cos(x)
yeah, but the derivative of function xsin(x) + cos(x) is not xcos(x)
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1 week ago
#16
(Original post by Melvin Guna)
yeah, but the derivative of function xsin(x) + cos(x) is not xcos(x)
What is it?
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1 week ago
#17
(Original post by mqb2766)
What is it?
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1 week ago
#18
(Original post by DFranklin)
You can have isolated points where the gradient is zero and still have a strictly increasing function. E.g. x^3 is strictly increasing (everywhere) but has derivative 0 at x = 0. (Most A-level references won't deal with this edge case, however).
thank you for clarifying
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1 week ago
#19
(Original post by the bear)
thank you for clarifying
No problem! FWIW, RDKGames is correct that the mathematical convention is that "increasing" includes the case. I confess I was somewhat surprised as "non-decreasing" seems a better (less ambiguous) way of dealing with this (and that was what I *thought* the convention was).
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