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    Hi, I was just wondering what I was doing wrong on this C3 question :confused: It's probably something quite simple aswell

    f:x\rightarrow 2(x+1) , x\in \mathbb{R}
    g:x\rightarrow x^2 -9, x\in \mathbb{R}

    (a) Express gf in terms of x and state its domain and range.

    I ended up getting: gf(x) = 4x^2 + 8x -5 With a range of: gf(x) \ge -5

    Thanks,
    Gra.
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    i reckon your gf in terms of x is fine, but then i would find the minimum, when the differential = 0 , i.e. when 8x+8=0, i.e. when x=-1, so gf(x)>= -1

    might be wrong though!
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    I agree with your gf(x).

    -5 is the y intercept, but not the minimum point.

    Try completing the square.
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    (Original post by AnonyMatt)
    I agree with your gf(x).

    -5 is the y intercept, but not the minimum point.

    Try completing the square.
    Oh of course! Duh. Silly me. It was a mental lapse, obviously.

    Thankyou
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    I now still can't do the last part, part c!

    Using these again:
    f:x\rightarrow 2(x+1) , x\in \mathbb{R}
    g:x\rightarrow x^2 -9, x\in \mathbb{R}

    (c) The equation gf(x) - 2f(x) = a where a is a constant, has no real roots. Show that a <-10

    Thanks again,
    Gra.
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    (Original post by born_to_be_different)
    I now still can't do the last part, part c!

    Using these again:
    f:x\rightarrow 2(x+1) , x\in \mathbb{R}
    g:x\rightarrow x^2 -9, x\in \mathbb{R}

    (c) The equation gf(x) - 2f(x) = a where a is a constant, has no real roots. Show that a <-10

    Thanks again,
    Gra.
    4x^2 + 8x - 5 - 4x - 4 = a
    4x^2 - 4x - (9+a) = 0

    What's the condition for a quadratic to have no real roots?
 
 
 
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