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    I'm stuck on edexcel Jan 07 question 2b and c...

    I don't even understand what the question is asking me to do.

    Click for the Jan 07 paper

    I can't write it out because I'm not that great at using latex.

    Thank you,
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    2b. What's the minimum value of the function?

    2c. What's the minimum value of the denominator of f(x)? Then use the fact that two positive numbers always divide to give a positive number.

    Spoiler:
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    If the minimum value is greater than 0 then whatever x is, x^2 + x + 1 will always be greater than 0
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    Complete the square for the numerator, that will tell you the range of the numerator (which will always be greater than 0.75, hence always greater than 0). Then C is just showing that since the numerator and denominator will always be positive, f(x) must be positive too...
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    b) put x^2+x+1/(x+2)^2 = 0
    so x^2+x+1 = 0

    as it is (x+2)^2, which cannot be negative because it is squared, therefore x^2+x+1 > 0

    c) is asking to show x^2+x+1/(x+2)^2 is positive - basically because the denominator is square, and the numerator has a square (so they are of the same order) and as they are both squared, it has to be positive
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    So can I differentiate for 2b and subs the value of x into the equation?

    Edit - Thanks for your replies. Rep coming your way in the next week or so.
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    no, just use x^2+x+1/(x+2)^2 = 0; x^2+x+1= 0 times (x+2)^2 therefore, x^2+x+1=0
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    2b complete the square to show, 3 over 4 is your minimum value. therefore curve > 0

    2c the denominator is a square, and we proved the numerator is positive, therefore f(x) > 0
 
 
 
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Updated: January 19, 2010
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