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    Hey guys having a bit of a problem on (ii) and (iii) of this question, I don't know how to go about it. I'll type out whole question

    The function f is defined by f(x) = sq root (mx+7) - 4, where x is equal to or greater than -7/m and m is a positive constant.

    (i) a sequence of transformations maps the curve y = sq root x to curve f(x). give details of transformations. <<<---- ok

    (ii) Explain how you can tell that f is one-one function and find expression for f^-1 (x). <<<---- can explain why it is 1:1 function but don't know how to find expression

    (iii) y=f(x) and y =f^-1 (x) do not meet. Explain how it can be deduced that neither curve meets the line y=x, and hence determine the set of possible values of m.

    Thanks
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    (Original post by silly_billy)
    Hey guys having a bit of a problem on (ii) and (iii) of this question, I don't know how to go about it. I'll type out whole question

    The function f is defined by f(x) = sq root (mx+7) - 4, where x is equal to or greater than -7/m and m is a positive constant.

    (i) a sequence of transformations maps the curve y = sq root x to curve f(x). give details of transformations. <<<---- ok

    (ii) Explain how you can tell that f is one-one function and find expression for f^-1 (x). <<<---- can explain why it is 1:1 function but don't know how to find expression

    (iii) y=f(x) and y =f^-1 (x) do not meet. Explain how it can be deduced that neither curve meets the line y=x, and hence determine the set of possible values of m.

    Thanks
    i) stretch by 1/m, shift 7 left, four down.

    ii) you could argue since gradient is always increasing, it's one to one.

    f(x) = √(mx+7) - 4
    mx + 7 = (f(x)+4)²
    f^-1x = ((x+4)² - 7)/m = (x² + 8x + 9)/m

    iii) finding the inverse is essentially swapping the x and y values round. hence if f(x) ≠ f^-1(x), y≠x

    second part.. give me a second.
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    (Original post by chewwy)
    i) stretch by 1/m, shift 7 left, four down.

    ii) you could argue since gradient is always increasing, it's one to one.

    f(x) = √(mx+7) - 4
    mx + 7 = (f(x)+4)²
    f^-1x = ((x+4)² - 7)/m = (x² + 8x + 9)/m

    iii) finding the inverse is essentially swapping the x and y values round. hence if f(x) ≠ f^-1(x), y≠x

    second part.. give me a second.
    thank you so much. can i take u into the exam please?
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    i remember this question from june. i didnt get part 3 then either
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    (Original post by Undry1)
    i remember this question from june. i didnt get part 3 then either
    damn and u do further maths
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    http://www.thestudentroom.co.uk/show...9&postcount=14

    thats someone's explanation of how they did it in the exam.
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    i don't understand

    ok, so let's find the range of values of m that do cause an intersection. then the other values of m won't. so:

    f(x) = x = f^-1(x)
    m = x+8 + 9/x

    now what? surely this means (since x+8 + 9/x has a range everywhere but 0) that they'll intersect if m is anything but 0.

    :confused:
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    (Original post by chewwy)
    i don't understand

    ok, so let's find the range of values of m that do cause an intersection. then the other values of m won't. so:

    f(x) = x = f^-1(x)
    m = x+8 + 9/x

    now what? surely this means (since x+8 + 9/x has a range everywhere but 0) that they'll intersect if m is anything but 0.

    :confused:
    what i think the guy was saying was:
    x = (x²+8x+9)/m
    xm = x²+8x+9
    0 = x²+(8-m)x+9

    using the discriminant to find values of m which cause the equation to have no roots:

    (8-m)²- 36 < 0
    m² - 16m +28 < 0
    (m-14)(m-2) < 0

    2 < m < 14
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    Thnks you guys for all ur help - u explained it really well, will have to remember how to do them
 
 
 
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