how to show a trigonometric function is increasing?

Original post by dxnixl

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.

Reply 2

dxnixl

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?

Reply 3

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?

Reply 4

dxnixl

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?

Reply 5

the bear

strictly speaking the gradient must be positive for an increasing function. if is zero then we say it is "stationary"

Reply 6

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.

Reply 7

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

Reply 8

dxnixl

Original post by RDKGames

Alright thanks :smile:

Reply 9

Original post by dxnixl

Thanks!

please double check your derivative, did you remember to use the product rule for xsinx :smile:

Reply 10

http://www.sunshinemaths.com/topics/calculus/increasing-and-decreasing-curves/

Original post by dxnixl

Thanks!

please double check your derivative, did you remember to use the product rule for xsinx :smile:

Reply 11

the bear

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

Reply 12

mqb2766

Original post by Melvin Guna

please double check your derivative, did you remember to use the product rule for xsinx :smile:

The function is
xsin(x) + cos(x)

Reply 13

DFranklin

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).

Reply 14

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) :smile:

Reply 15

mqb2766

Original post by Melvin Guna

yeah, but the derivative of function xsin(x) + cos(x) is not xcos(x) :smile:

What is it?

Reply 16