Turn on thread page Beta
    • Thread Starter
    Offline

    21
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

    Name:  image.jpg
Views: 152
Size:  500.2 KB
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

    Name:  image.jpg
Views: 152
Size:  500.2 KB
    Lovely work on part(a), it's all correct!

    For part(b) you know that stationary points are solutions to the equation \frac{\mathrm{d}y}{\mathrm{d}x} = 0, can you solve this equation and find the x-coordinates of the stationary points?
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    Lovely work on part(a), it's all correct!

    For part(b) you know that stationary points are solutions to the equation \frac{\mathrm{d}y}{\mathrm{d}x} = 0, can you solve this equation and find the x-coordinates of the stationary points?
    So would I factorise the answer to part a to get 3(x+2)(x+2) and then solve for x?
    • TSR Support Team
    Offline

    21
    ReputationRep:
    TSR Support Team
    (Original post by homeland.lsw)
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

    Name:  image.jpg
Views: 152
Size:  500.2 KB
    For 7.b., you need to remember the important fact that \frac{dy}{dx}=0 at stationary points and that \frac{d^2y}{dx^2}\leq 0 at maxima (i.e. there is downwards curvature, the gradient is decreasing) and \frac{d^2y}{dx^2}\geq 0 at minima (i.e. there is upwards curvature, the gradient is increasing).

    This should help you answer question 8.b. too. For 8.c, think about the shape of inflection points - how might you show it's an inflection point? Hint: you may want to use \frac{dy}{dx}.
    Offline

    4
    ReputationRep:
    (Original post by homeland.lsw)
    Hey, I was just confused on how to actually do these two questions.
    I did number 7a (I got 3x^2 + 12x +12) but not the other three. I would appreciate some help. Thanks! 😀

    Name:  image.jpg
Views: 152
Size:  500.2 KB
    For 7b) Derive the equation again, as it is a stationary point d2y/dx2 must =0, rearrange for x and sub it back into the orignal equation to get they y coordinate
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    So would I factorise the answer to part a to get 3(x+2)(x+2) and then solve for x?
    Yes, that's correct. Once you've found the requisite values of x you then need to check whether \frac{d^2y}{dx^2} is positive, negative or zero for each of the x's that you've found to categorise it as a minimum, maximum or inflection.
    Offline

    22
    ReputationRep:
    (Original post by K-Fox)
    For 7b) Derive the equation again,
    Do you mean differentiate?
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    Yes, that's correct. Once you've found the requisite values of x you then need to check whether \frac{d^2y}{dx^2} is positive, negative or zero for each of the x's that you've found to categorise it as a minimum, maximum or inflection.
    Ok great, so my answer is this...
    Attachment 511841511843511900

    Also what's the story with that \frac{d^2y}{dx^2}?It's a bit confusing...

    Edit I don't know why there are three attachments
    Attached Images
       
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    Ok great, so my answer is this...
    Attachment 511841511843511900

    Also what's the story with that \frac{d^2y}{dx^2}?It's a bit confusing...

    Edit I don't know why there are three attachments
    \frac{d^2y}{dx^2}= \frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (3x^2 + 12x +12) have you seen this before? It's the derivative of the derivative.
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    \frac{d^2y}{dx^2}= \frac{d}{dx} \frac{dy}{dx} = \frac{d}{dx} (3x^2 + 12x +12) have you seen this before? It's the derivative of the derivative.
    I haven't no...we covered maxima and minima using the method from question 7 and a table...

    Anyway my question 7b is correct I presume? And I'll get on with question 8.
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    Ok great, so my answer is this...
    Oh, and your stationary points is perfect. Good work! You just need to determine whether it's a minimum, maximum or inflection point now.

    Once you find your second derivative (by differentiating your first derivative) you should get an equation that when you plug x=-2 in, it'll either give you a positive value (x=-2 is a minimum), a negative value (x=-2 is a maximum) or a zero (x=-2 is an inflection point).
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    Oh, and your stationary points is perfect. Good work! You just need to determine whether it's a minimum, maximum or inflection point now.

    Once you find your second derivative (by differentiating your first derivative) you should get an equation that when you plug x=-2 in, it'll either give you a positive value (x=-2 is a minimum), a negative value (x=-2 is a maximum) or a zero (x=-2 is an inflection point).
    And if by magic it gave me zero!!!

    Name:  image.jpg
Views: 110
Size:  518.8 KB

    Also you remind me of my maths teacher by saying "plug in" :lol:
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    And if by magic it gave me zero!!!

    Name:  image.jpg
Views: 110
Size:  518.8 KB

    Also you remind me of my maths teacher by saying "plug in" :lol:
    Yeees! Inflection it is. I get the same thing.

    I really should be saying "substitute" instead of "plug in". :lol:
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    Yeees! Inflection it is. I get the same thing.

    I really should be saying "substitute" instead of "plug in". :lol:
    Ok so this is my number 8...
    Name:  image.jpg
Views: 59
Size:  518.5 KB

    Attachment 511857511859
    Attached Images
     
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    Ok so this is my number 8...
    Attachment 511857511859
    All correct. :yep:
    • Thread Starter
    Offline

    21
    (Original post by Zacken)
    All correct. :yep:
    I'm practically a mathematician now!! :lol:
    Offline

    22
    ReputationRep:
    (Original post by homeland.lsw)
    I'm practically a mathematician now!! :lol:
    I concur.
    Offline

    22
    ReputationRep:
    (Original post by longshot100)
    ...
    He's gotten his answer already.
    Offline

    11
    ReputationRep:
    (Original post by Zacken)
    He's gotten his answer already.
    Oh lol, sorry
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: March 10, 2016

985

students online now

800,000+

Exam discussions

Find your exam discussion here

Poll
Should predicted grades be removed from the uni application process
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.