Turn on thread page Beta
    • Thread Starter
    Offline

    16
    ReputationRep:
    The question is:  f(x) = x/x^2+2 Find the set of values of x for which f'(x)<0
    • TSR Support Team
    Offline

    21
    ReputationRep:
    TSR Support Team
    (Original post by AxSirlotl)
    The question is:  f(x) = x/x^2+2 Find the set of values of x for which f'(x)<0
    The first step is obviously to get an expression for f'(x), have you managed to do that?
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by Plagioclase)
    The first step is obviously to get an expression for f'(x), have you managed to do that?
    yeah boii
    •  Official Rep
    Offline

    12
    ReputationRep:
     Official Rep
    (Original post by AxSirlotl)
    The question is:  f(x) = x/x^2+2 Find the set of values of x for which f'(x)<0
    You have to use the quotient rule and then look at the numerator.

    Could you show your workings?
    • TSR Support Team
    Offline

    21
    ReputationRep:
    TSR Support Team
    (Original post by AxSirlotl)
    yeah boii
    Well then, you've got your inequality, you just need to solve that.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by BrasenoseAdm)
    You have to use the quotient rule and then look at the numerator.

    Could you show your workings?
    I got  f'(x) = -2x^2/x^2+2
    Sorry, I can't do proper fractions because I don't understand the guide heh
    • Community Assistant
    Online

    18
    ReputationRep:
    Community Assistant
    (Original post by AxSirlotl)
    I got  f'(x) = -2x^2/x^2+2
    Sorry, I can't do proper fractions because I don't understand the guide heh
    Your derivative is wrong. To write a fraction it's \frac{numerator}{denominator}, but the notation you're using is acceptable anyway.
    •  Official Rep
    Offline

    12
    ReputationRep:
     Official Rep
    (Original post by NotNotBatman)
    Your derivative is wrong. To write a fraction it's \frac{numerator}{denominator}, but the notation you're using is acceptable anyway.
    Does TSR detect LaTex automatically? We tried replying with the coding:

    \frac{x}{x^2+2}

    &

    [tex]\frac{x}{x^2+2}
    • Community Assistant
    Online

    18
    ReputationRep:
    Community Assistant
    (Original post by BrasenoseAdm)
    Does TSR detect LaTex automatically? We tried replying with the coding:

    \frac{x}{x^2+2}

    &

    [tex]\frac{x}{x^2+2}
    [tex] maths things go here [/ tex] , but without the space after the /
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by NotNotBatman)
    Your derivative is wrong. To write a fraction it's \frac{numerator}{denominator}, but the notation you're using is acceptable anyway.
    I keep getting the same answer ;-;

    I tried doing fractions but it kept putting my numerator in front of my fraction and leaving the top of the fraction blank, so I can't really show my working very easily.
    •  Official Rep
    Offline

    12
    ReputationRep:
     Official Rep
    (Original post by AxSirlotl)
    I keep getting the same answer ;-;

    I tried doing fractions but it kept putting my numerator in front of my fraction and leaving the top of the fraction blank, so I can't really show my working very easily.
    Thanks for the LaTex tip!

    Problem: differentiate \frac{x}{x^2+2}

    Technique: quotient rule \frac{vu' - uv'}{v^2}

    u = x and v = X^2+2, u' = derivative of u and v'= derivative of v
    • Community Assistant
    Offline

    17
    ReputationRep:
    Community Assistant
    Use the chain rule for differentiation:

    h(x) = f(x)/g(x)

    h'(x) = f'(x)g(x)-f(x)g'(x) / [g(x)]^2

    The last step is trivial; let the inequality be less than 0 and solve it accordingly.
    Offline

    17
    ReputationRep:
    Is that  \frac{x}{x^2} + 2 or  \frac{x}{x^2+2}
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by zeldor711)
    Is that  \frac{x}{x^2} + 2 or  \frac{x}{x^2+2}
    The latter
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by BrasenoseAdm)
    Thanks for the LaTex tip!

    Problem: differentiate \frac{x}{x^2+2}

    Technique: quotient rule \frac{vu' - uv'}{v^2}

    u = x and v = X^2+2, u' = derivative of u and v'= derivative of v
    (Original post by thekidwhogames)
    Use the chain rule for differentiation:

    h(x) = f(x)/g(x)

    h'(x) = f'(x)g(x)-f(x)g'(x) / [g(x)]^2

    The last step is trivial; let the inequality be less than 0 and solve it accordingly.
    I got my derivative as -1/(x^2+2), is that right ;-;
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by AxSirlotl)
    I got my derivative as -1/(x^2+2), is that right ;-;
    Not quite.

    If you post your working we would just be able to spot where you're going wrong.
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by RDKGames)
    Not quite.

    If you post your working we would just be able to spot where you're going wrong.
    Using the quotient rule:

    u = x v = x^2 + 2

    u' = 1 v' = 2x

    Then I got
    numerator: (x^2 + 2) - 2x^2

    denominator: (x^2 + 2)^2

    So on the top I had v x u' - u x v' all over v^2
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by AxSirlotl)
    Using the quotient rule:

    u = x v = x^2 + 2

    u' = 1 v' = 2x

    Then I got
    numerator: (x^2 + 2) - 2x^2

    denominator: (x^2 + 2)^2

    So on the top I had v x u' - u x v' all over v^2
    OK, so how did you go from \frac{2-x^2}{(x^2+2)^2} to -\frac{1}{x^2+2}?
    • Thread Starter
    Offline

    16
    ReputationRep:
    (Original post by RDKGames)
    OK, so how did you go from \frac{2-x^2}{(x^2+2)^2} to -\frac{1}{x^2+2}?
    I've just realised that's wrong, I mean I've had another go at it and I'm back to (-x^2 + 2)/(x^2+2)^2
    I'm not sure where to go from there, if I need make that into an inequality then I'm not sure what to do after that.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by AxSirlotl)
    I've just realised that's wrong, I mean I've had another go at it and I'm back to (-x^2 + 2)/(x^2+2)^2
    I'm not sure where to go from there, if I need make that into an inequality then I'm not sure what to do after that.
    (x^2+2)^2 &gt; 0 so \frac{2-x^2}{(x^2+2)^2} &lt;0 \Rightarrow 2-x^2 &lt;0
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 5, 2017

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
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.