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    The equation f(t)=ke^t, where k>0, has one real solution.
    Find the value of k.

    I tried working it out and couldn't do it. Then I plotted it on a graph and saw that the only time there was a root (as far as I could tell), was when k=0 which had infinite roots as well as k not being positive.
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    (Original post by KloppOClock)
    The equation f(t)=ke^t, where k>0, has one real solution.
    Find the value of k.

    I tried working it out and couldn't do it. Then I plotted it on a graph and saw that the only time there was a root (as far as I could tell), was when k=0 which had infinite roots as well as k not being positive.
    As it stands you just have a function. You need to equate it to something to get an equation that can be solved. You need to check the exact wording of your original question. Or post a link/photo.
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    (Original post by ghostwalker)
    As it stands you just have a function. You need to equate it to something to get an equation that can be solved. You need to check the exact wording of your original question. Or post a link/photo.
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    (Original post by KloppOClock)
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    This is definitely the whole question? There aren't any earlier parts where f(t) is defined? Because that question doesn't make any sense- you need more information.
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    (Original post by sindyscape62)
    This is definitely the whole question? There aren't any earlier parts where f(t) is defined? Because that question doesn't make any sense- you need more information.
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    (Original post by KloppOClock)
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    Ok, now it's solvable- do you see why? Part c puts f(t) equal to something else, so if you don't know what f(t) is (defined at the start) you can't get anywhere with it.

    Your first step is to write  3-2e^-^t=ke^t
    Then solve it for t. Try multiplying through by  e^t and you should get a quadratic. Quadratics either produce 0, 1 or 2 different real roots- think about the condition for a quadratic to produce just one. This will give you a solvable equation in terms of k.
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    (Original post by sindyscape62)
    Ok, now it's solvable- do you see why? Part c puts f(t) equal to something else, so if you don't know what f(t) is (defined at the start) you can't get anywhere with it.

    Your first step is to write  3-2e^-^t=ke^t
    Then solve it for t. Try multiplying through by  e^t and you should get a quadratic. Quadratics either produce 0, 1 or 2 different real roots- think about the condition for a quadratic to produce just one. This will give you a solvable equation in terms of k.
    oh right i forgot about the original fx bit, thanks
 
 
 
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