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Thread: graphs -- urgent

  1. #1
    Jul 2007

    graphs -- urgent

    it'd be great if anyone could help me on this, im kinda stuck..

    Okay, so there's a diagram, and it shows the graphs of f(x) = 1 + e^2x and g(x) = 10x + 2, where 0 [FONT='Times New Roman','serif']≤ x [FONT='Times New Roman','serif']≤ 1.5[/FONT][/FONT]
    [FONT='Times New Roman','serif'][/FONT]
    (a) (i) write down an expression for the vertical distance p between the graphs of f and g.
    (ii) given that p has a maximum value for 0 [FONT='Times New Roman','serif']≤ x [FONT='Times New Roman','serif']≤ 1.5, find the value of x at which this occurs. [/FONT][/FONT]

    (b) (i) i dont know if you guys could help me out on this one as well without the diagram, but here it is. so the graph of y = f(x) is there, and it says that when x = a and y = 5, what is f^-1(x).
    (ii) hence show that a - ln2

    thanks a lot!
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  2. #2
    MHF Contributor red_dog's Avatar
    Jun 2007
    Medgidia, Romania
    The distance between the graphs is p(x)=10x-e^{2x}+1
    p'(x)=10-2e^{2x}. Solving the equation p'(x)=0 we get \displaystyle x=\frac{1}{2}\ln 5 which is a point of maximum.
    Then the maximum of p is p\left(\frac{1}{2}\ln 5\right)=5\ln 5-4.

    To find f^{-1}, from the equality f(x)=y we must represent x as a fucntion of y.
    So f(x)=y\Rightarrow 1+e^{2x}=y\Rightarrow e^{2x}=y-1\Rightarrow x=\frac{1}{2}\ln (y-1).
    Then \displaystyle f^{-1}(x)=\frac{1}{2}\ln (x-1)

    f(a)=5\Leftrightarrow 1+e^{2a}=5\Leftrightarrow e^{2a}=4\Leftrightarrow 2a=2\ln 2\Leftrightarrow a=\ln 2
    Attached Thumbnails Attached Thumbnails graphs -- urgent-graph2.jpg  
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