how to show a trigonometric function is increasing?

Watch
dxnixl
Badges: 17
Rep:
?
#1
Report Thread starter 1 week ago
#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.

Name:  08412422-64BE-4A26-8110-666A7E168B1D.jpeg
Views: 5
Size:  19.5 KB
Name:  C77EB7F3-2402-4649-95C3-2053825253ED.jpeg
Views: 5
Size:  125.0 KB

Thanks!
Last edited by dxnixl; 1 week ago
0
reply
RDKGames
Badges: 20
Rep:
?
#2
Report 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 \geq 0. So as long as you can reason out why the derivative never negative, then you're good.
0
reply
dxnixl
Badges: 17
Rep:
?
#3
Report Thread starter 1 week ago
#3
(Original post by RDKGames)
A function is increasing if its derivative is \geq 0. 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?
0
reply
RDKGames
Badges: 20
Rep:
?
#4
Report 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 \geq 0.

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?
0
reply
dxnixl
Badges: 17
Rep:
?
#5
Report Thread starter 1 week ago
#5
(Original post by RDKGames)
Not again ... you already got the derivative, you just need to explain why it is \geq 0.

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?
0
reply
the bear
Badges: 20
Rep:
?
#6
Report 1 week ago
#6
strictly speaking the gradient must be positive for an increasing function. if is zero then we say it is "stationary"
0
reply
RDKGames
Badges: 20
Rep:
?
#7
Report 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.
0
reply
RDKGames
Badges: 20
Rep:
?
#8
Report 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 a_{n+1} \geq a_n

{ 0 ,1 , 2 , ... } is strictly increasing because a_{n+1} > a_n
0
reply
dxnixl
Badges: 17
Rep:
?
#9
Report Thread starter 1 week ago
#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
0
reply
Melvin Guna
Badges: 5
Rep:
?
#10
Report 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.

Name:  08412422-64BE-4A26-8110-666A7E168B1D.jpeg
Views: 5
Size:  19.5 KB
Name:  C77EB7F3-2402-4649-95C3-2053825253ED.jpeg
Views: 5
Size:  125.0 KB

Thanks!
please double check your derivative, did you remember to use the product rule for xsinx
0
reply
Melvin Guna
Badges: 5
Rep:
?
#11
Report 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.

Name:  08412422-64BE-4A26-8110-666A7E168B1D.jpeg
Views: 5
Size:  19.5 KB
Name:  C77EB7F3-2402-4649-95C3-2053825253ED.jpeg
Views: 5
Size:  125.0 KB

Thanks!
please double check your derivative, did you remember to use the product rule for xsinx
0
reply
the bear
Badges: 20
Rep:
?
#12
Report 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 a_{n+1} \geq a_n

{ 0 ,1 , 2 , ... } is strictly increasing because a_{n+1} > a_n
http://www.sunshinemaths.com/topics/...easing-curves/
0
reply
mqb2766
Badges: 18
Rep:
?
#13
Report 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)
0
reply
DFranklin
Badges: 18
Rep:
?
#14
Report 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).
0
reply
Melvin Guna
Badges: 5
Rep:
?
#15
Report 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)
0
reply
mqb2766
Badges: 18
Rep:
?
#16
Report 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?
0
reply
Melvin Guna
Badges: 5
Rep:
?
#17
Report 1 week ago
#17
(Original post by mqb2766)
What is it?
apologies ... I misread your earlier comment-you are right
0
reply
the bear
Badges: 20
Rep:
?
#18
Report 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
0
reply
DFranklin
Badges: 18
Rep:
?
#19
Report 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 \leq 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).
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

With no certainty that exams next year wil take place, how does this make you feel?

More motivated (29)
26.61%
Less motivated (80)
73.39%

Watched Threads

View All
Latest
My Feed