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    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
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    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.
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    (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?
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    -3e^3x -2x
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    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.
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    (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)
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    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...
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    (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))
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    Keep going...
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    -3(e^(3x)-x(2)) < 0

    e^(3x)-x(2) < 0
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    Being pedantic, a function is decreasing when  f'(x) \leq 0 , whereas it is strictly decreasing when  f'(x) &lt; 0 ...
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    (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.
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    Oh! yea anything -x^2 becomes positive, right
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    (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)?
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    It's an exponential can never be negative
 
 
 
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