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Math Help - A level maths 06

  1. #1
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    Talking A level maths 06

    some more ...i have so many things to solve but i dont know where to start from ,so urgent help needed!!!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by carlasader View Post
    some more ...i have so many things to solve but i dont know where to start from ,so urgent help needed!!!
    us drawing the diagram for you won't help, since we'd use a graphing utility and you won't know what happens. if you have any specific questions about graphing, just say so and i'll try to answer them. otherwise i'll find a from question 2

    y = (x - 1)^2 (x + a)

    Now we are told that the graph passes through (2,5), this means when x = 2, y = 5

    \Rightarrow 5 = (2 - 1)^2 (2 + a)

    \Rightarrow 5 = 2 + a

    \Rightarrow a = 3

    Hopefully you can take it from here. if not, ask SPECIFIC questions about graphing
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by carlasader View Post
    some more ...i have so many things to solve but i dont know where to start from ,so urgent help needed!!!
    here's 3


    Let h(x) = x^3
    Let f(x) = x - 1
    Let g(x) = 2x

    Then we can form y = 2(x - 1)^3 via a composite function of f, g and h

    Let y = (g \circ h \circ f )(x)

    \Rightarrow y = g(h(f(x)))

    \Rightarrow y = g((x - 1)^3)

    \Rightarrow y = 2(x - 1)^3


    As i said, graphing for you won't help, so let's move on to finding the inverse of y

    y = 2(x - 1)^3

    For inverse, we solve x = 2(y - 1)^3 for y

    \Rightarrow \frac {x}{2} = (y - 1)^3

    \Rightarrow \sqrt [3] { \frac {x}{2}} = y - 1

    \Rightarrow y = \sqrt [3] { \frac {x}{2}} - 1 is the inverse function

    As for the graphical relationship. we obtain the graph of the inverse of a function by reflecting the original function in the line y = x
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