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Thread: 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

    $\displaystyle y = (x - 1)^2 (x + a)$

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

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

    $\displaystyle \Rightarrow 5 = 2 + a$

    $\displaystyle \Rightarrow a = 3$

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

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

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

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

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

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


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

    $\displaystyle y = 2(x - 1)^3$

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

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

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

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