- Forums
###### Stationary point for curve y=3sec(2x)

Watch

2 months ago

I recently learnt gradients to non-algebraic functions and now realising some are available on the formula sheet.

Anyway, so if I know the gradient is

F’(x)=6sec(2x)tan(2x) and stationary point being f’(x)=0

Is there a shortcut to find the value of x? I have tried to use the double angle identities with no fruition as I’m left with both sin and cos still in the function but I might be over looking something.

P.S: how do you upload photos now? I don’t see an option on the phone.

Anyway, so if I know the gradient is

F’(x)=6sec(2x)tan(2x) and stationary point being f’(x)=0

Is there a shortcut to find the value of x? I have tried to use the double angle identities with no fruition as I’m left with both sin and cos still in the function but I might be over looking something.

P.S: how do you upload photos now? I don’t see an option on the phone.

(edited 2 months ago)

Original post by KingRich

I recently learnt gradients to non-algebraic functions and now realising some are available on the formula sheet.

Anyway, so if I know the gradient is

F’(x)=6sec(2x)tan(2x) and stationary point being f’(x)=0

Is there a shortcut to find the value of x? I have tried to use the double angle identities with no fruition as I’m left with both sin and cos still in the function but I might be over looking something.

Anyway, so if I know the gradient is

F’(x)=6sec(2x)tan(2x) and stationary point being f’(x)=0

Is there a shortcut to find the value of x? I have tried to use the double angle identities with no fruition as I’m left with both sin and cos still in the function but I might be over looking something.

The gradient should be 2sec(2x)tan(2x). Not important for stationary points as its when its = 0 but ... So just think about factors so when sec(2x) or tan(2x) are zero as their product must be zero.

If unsure, a sketch of sec(2x) might be helpful for thinking about it.

No option for uploading photos at the moment after the recent "upgrade". Youd need to upload elsewhere and link with the picture icon/button.

(edited 2 months ago)

Original post by mqb2766

The gradient should be 2sec(2x)tan(2x). Not important for stationary points as its when its = 0 but ... So just think about factors so when sec(2x) or tan(2x) are zero as their product must be zero.

If unsure, a sketch of sec(2x) might be helpful for thinking about it.

No option for uploading photos at the moment after the recent "upgrade". Youd need to upload elsewhere and link with the picture icon/button.

If unsure, a sketch of sec(2x) might be helpful for thinking about it.

No option for uploading photos at the moment after the recent "upgrade". Youd need to upload elsewhere and link with the picture icon/button.

Apologise, I missed the 3 from the original function.

Well that sucks about the photo issue.

ah, so without recalling the shape of the sec(x) graph, it’s sorta impossible to answer.

Original post by KingRich

Apologise, I missed the 3 from the original function.

Well that sucks about the photo issue.

ah, so without recalling the shape of the sec(x) graph, it’s sorta impossible to answer.

Well that sucks about the photo issue.

ah, so without recalling the shape of the sec(x) graph, it’s sorta impossible to answer.

If it was 3sec(2x) then the derivative is correct. To find the stationary points you could note sec=1/cos so this corresponds to

sec(2x)tan(2x) = sin(2x)/cos^2(2x) = 0

and you can argue about either expression being zero to get the stationary points. However, you should be able to sketch sec, even if you have to use the 1/cos definition

Original post by mqb2766

If it was 3sec(2x) then the derivative is correct. To find the stationary points you could note sec=1/cos so this corresponds to

sec(2x)tan(2x) = sin(2x)/cos^2(2x) = 0

and you can argue about either expression being zero to get the stationary points. However, you should be able to sketch sec, even if you have to use the 1/cos definition

sec(2x)tan(2x) = sin(2x)/cos^2(2x) = 0

and you can argue about either expression being zero to get the stationary points. However, you should be able to sketch sec, even if you have to use the 1/cos definition

Let’s say I could recalled sec x, then sec(2x)=0 would be invalid as it doesn’t cross the x-axis..

Tan(x)= x+π

Tan(2x) would suggest a stretch in x… okay, so if 0<x<π

Then new range would be 0<x<2π

Am I on the right path?

Original post by KingRich

Let’s say I could recalled sec x, then sec(2x)=0 would be invalid as it doesn’t cross the x-axis..

Tan(x)= x+π

Tan(2x) would suggest a stretch in x… okay, so if 0<x<π

Then new range would be 0<x<2π

Am I on the right path?

Tan(x)= x+π

Tan(2x) would suggest a stretch in x… okay, so if 0<x<π

Then new range would be 0<x<2π

Am I on the right path?

Agree sec is <=-1 or >=1 so goes nowhere near the x-axis and, from a sketch/definition, the points where it has stationary points must correspond to stationary points of cos.

But youre sort of "right" in that the stationary points of sec must correspond to where tan is zero.

tan(x) = 0

when x = k*pi, where k is an integer. So 0,pi,-pi,2pi,-2pi,... Then just adapt when the argument is 2x.

You could argue from the second epxression that its when sin(2x)=0 as well as obv the two answers must be the same.

So, the of chance I can recall sec 2x, I can take the long path using identities and consider that sin x = x+π

Hence, x=0,π and 2π

So,

2x=0

2x=π

2x= 2π

So, x=2/π can be the only permissible x for the one point it’s looking for.

Then inputting that into the original equation tells us y=-3

Hence, x=0,π and 2π

So,

2x=0

2x=π

2x= 2π

So, x=2/π can be the only permissible x for the one point it’s looking for.

Then inputting that into the original equation tells us y=-3

Original post by KingRich

So, the of chance I can recall sec 2x, I can take the long path using identities and consider that sin x = x+π

Hence, x=0,π and 2π

So,

2x=0

2x=π

2x= 2π

So, x=2/π can be the only permissible x for the one point it’s looking for.

Then inputting that into the original equation tells us y=-3

Hence, x=0,π and 2π

So,

2x=0

2x=π

2x= 2π

So, x=2/π can be the only permissible x for the one point it’s looking for.

Then inputting that into the original equation tells us y=-3

Not sure whether its typos or genuine mistakes but writing

sin(x) = x + pi

makes no sense in this context, and similarly for the previous post. The solutions correspond to where sin or tan is zero, so

sin(2x) = 0

corresponds to

2x = k*pi

where k is an integer. Dividing by 2 gives

x = k*pi/2

and depending on the domain given in the question (not posted) its includes 0,+/-pi/2,+/-pi, ... The way youve written it up is somewhere between confusing and wrong (2/pi?)?

- Sketching cubic
- as level ocr maths question - help
- Second Year A level maths differentiation
- Can anyone solve this equation?
- Stationary points!! Please HELP!!!!!
- Maths
- Point of inflection
- maths question help!!
- Differentiation
- A-level Maths, Integration, Finding area between curves & Lines.
- differentiation stationary points
- Countour plots
- Laws of logarithms
- equations of tangents and normals
- Quadratic
- If curve y=f(x) has a max point at x=a, show that y=1/f(x) has a min point at x=a
- The third derivative
- Maths Q parametric equations
- When transforming a graph, does it matter which order you list the transformations?
- Finding stationary point with a double angle identity

- Will i be allowed to change my a-level options after i get an offer/ results back?
- Functions maths question
- Phoenix Group Rotation Program
- Economics A-Level Pros and Cons?
- NICS Staff Officer and Deputy Principal recruitment 2022 2023
- Korean Studies Sheffield
- Official: Imperial College London A100 2024 Entry Applicants
- where to study
- I got A*AA for my A-Levels - ask me anything.
- Can I do engineering without chemistry in a levels
- "Can't you hear me, S.O.S.?": TLG's PhD study blog!
- Forum Game Delay
- Finance Declined
- Oxford Environmental Research DTP
- UCL Medicine A100 2024
- Official: University of East Anglia (UEA) A104 (Gateway Year) 2024 Entry
- September 2024
- Girls Vs. Boys (Part 46)
- I'm almost 21 and have never been in a relationship
- A level physics - Engineering question (5)

- Year 9-2023-24
- TSR Community Awards 2023: Most Helpful Thread of the year - VOTING NOW OPEN
- Oxford Creative Writing MSt 2024
- Korean/Japanese Studies 2024 Applicants
- Official University of Edinburgh Applicant Thread for 2024
- ੈ✩‧₊˚ year 13 academic comeback
- Scientist Training Programme 2024 Applicants
- Official: Brunel University A100 2024 Entry Applicants
- Philosophy/Politics Durham offer?
- Official London School of Economics and Political Science 2024 Applicant Thread
- What after a level 4 apprenticeship
- Cambridge postgraduate application: Decision pending - application is under review
- The Student Room Update: New Look, New Homepage and More!
- Paper rounds
- Official Veterinary Medicine Applicants thread 2024 entry
- GCSE disqualification
- [Official thread] Hamas-Israel Conflict
- Bank of America apprenticeship 2023
- UCL LAW 2024 LNAT cut off
- Official KCL Offer Holders Thread for 2024 entry

- UKMT Intermediate Math Challenge 2024 - Discussion
- Help with complex summation further maths a levels
- GCSE Mathematics Study Group 2023-2024
- Could I have some help with this suvat question?
- MAT practice
- A-level Mathematics Study Group 2023-2024
- Alevel Maths Question
- weird cosine question
- Series function not differentiable at a point
- hyperbolic function catenary problem

- Mock set 4 paper 2 q14 a level maths (4 distinct points)
- can anyone answer this A level vectors question (very challenging)
- STEP foundation module help pls
- Ukmt IMC 2024
- ukmt imc
- This maths question is driving me crazy
- Maths Mechanics
- Confused on surds question
- HNC MATHS A2 Task 3 (Radio Transmitters)
- Senior Maths Challenge 2023

- UKMT Intermediate Math Challenge 2024 - Discussion
- Help with complex summation further maths a levels
- GCSE Mathematics Study Group 2023-2024
- Could I have some help with this suvat question?
- MAT practice
- A-level Mathematics Study Group 2023-2024
- Alevel Maths Question
- weird cosine question
- Series function not differentiable at a point
- hyperbolic function catenary problem

- Mock set 4 paper 2 q14 a level maths (4 distinct points)
- can anyone answer this A level vectors question (very challenging)
- STEP foundation module help pls
- Ukmt IMC 2024
- ukmt imc
- This maths question is driving me crazy
- Maths Mechanics
- Confused on surds question
- HNC MATHS A2 Task 3 (Radio Transmitters)
- Senior Maths Challenge 2023

The Student Room and The Uni Guide are both part of The Student Room Group.

© Copyright The Student Room 2024 all rights reserved

The Student Room and The Uni Guide are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: Imperial House, 2nd Floor, 40-42 Queens Road, Brighton, East Sussex, BN1 3XB