Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    0
    ReputationRep:
    just a quick question

    Function f is defined for all real values of x by
    f(X) = 2-e^3x-x^2

    prove that it's decreasing function
    Offline

    15
    ReputationRep:
    Well, this is a tough one if it's C3.

    First of all, you need to work out f'(x). A function is decreasing when f'(x) < 0 for some x. Here, you're aked to prove it for all x, so you need to show that f'(x) is ALWAYS negative. It's not trivial to do - try doing an appropriate sketch. Post back with what you think f'(x) is and what you think I'm asking you to sketch.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Swayum)
    Well, this is a tough one if it's C3.

    First of all, you need to work out f'(x). A function is decreasing when f'(x) < 0 for some x. Here, you're aked to prove it for all x, so you need to show that f'(x) is ALWAYS negative. It's not trivial to do - try doing an appropriate sketch. Post back with what you think f'(x) is and what you think I'm asking you to sketch.
    erm.. I think f'(X) would be -3e^3x-3x^2 is that right?

    Your the only one who can handle this, half an hour no one answered thanks... what shall I do after that?
    Offline

    2
    ReputationRep:
    -3e^3x -2x
    Offline

    15
    ReputationRep:
    No, f'(x) = -3e^(3x) - 2x...

    In fact, I think the question is wrong or you've written it incorrectly.

    We're trying to prove that -3e^(3x) - 2x < 0 for all real x because this shows f(x) is decreasing, yes?

    But there's a counterexample to that: x = -3, then -3e^(-9) + 6 > 0, so f(x) is not always decreasing. Hell, there's an x so that f'(x) = 0 too.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Swayum)
    No, f'(x) = -3e^(3x) - 2x...

    In fact, I think the question is wrong or you've written it incorrectly.

    We're trying to prove that -3e^(3x) - 2x < 0 for all real x because this shows f(x) is decreasing, yes?

    But there's a counterexample to that: x = -3, then -3e^(-9) + 6 > 0, so f(x) is not always decreasing. Hell, there's an x so that f'(x) = 0 too.
    I am ever so sorry yes, I have written it wrong, hence I differentiated wrong... the question is f(x) = 2-e^(3x)-x(3)
    Offline

    15
    ReputationRep:
    Ok, that actually makes it easier than I was thinking.

    So f'(x) = -3e^(3x) - 3x^2

    We need to prove that f'(x) < 0 for all x, i.e. that -3e^(3x) - 3x^2 < 0 for all x.

    Can you do this? Start by rearranging to something nicer...
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Swayum)
    Ok, that actually makes it easier than I was thinking.

    So f'(x) = -3e^(3x) - 3x^2

    We need to prove that f'(x) < 0 for all x, i.e. that -3e^(3x) - 3x^2 < 0 for all x.

    Can you do this? Start by rearranging to something nicer...
    would we factorise it like this:

    -3(e^(3x)-x(2))
    Offline

    15
    ReputationRep:
    Keep going...
    • Thread Starter
    Offline

    0
    ReputationRep:
    -3(e^(3x)-x(2)) < 0

    e^(3x)-x(2) < 0
    Offline

    12
    ReputationRep:
    Being pedantic, a function is decreasing when  f'(x) \leq 0 , whereas it is strictly decreasing when  f'(x) &lt; 0 ...
    Offline

    15
    ReputationRep:
    (Original post by Unbounded)
    Being pedantic, a function is decreasing when  f'(x) \leq 0 , whereas it is strictly decreasing when  f'(x) &lt; 0 ...
    Yeah, I'd written that at first actually, then thought it's the other way round. Meh, long time since I've done any of this :p:. It's not as pedantic as you might think.

    (Original post by asem93)
    -3(e^(3x)-x(2)) < 0

    e^(3x)-x(2) < 0
    So e^3x >= -x^2 is what we're trying to prove. Can you think of why this must be true just looking at the equation?

    If not, sketch y = e^3x and y = -x^2 on the same graph.
    • Thread Starter
    Offline

    0
    ReputationRep:
    Oh! yea anything -x^2 becomes positive, right
    Offline

    15
    ReputationRep:
    (Original post by asem93)
    Oh! yea anything -x^2 becomes positive, right
    x^2 is always positive, but -x^2 is always negative for real x. What about e^(3x)?
    • Thread Starter
    Offline

    0
    ReputationRep:
    It's an exponential can never be negative
 
 
 
  • 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.

  • Poll
    Have you ever been hacked?
    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
  • 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.

  • 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.