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    (Original post by jessyjellytot14)
    "Prove, by counter-example, that the statement
    “ln (3x 2 + 5x + 3) ≥ 0 for all real values of x” is false."

    Hmmm I'm not quite sure how to go about this so i'll post my working below & could you check it please?

    Also can some explain what a real number is? I don't do further maths but I'm assuming it's just a whole integer.. So for example, -2/3 isn't a real number...?

    Thanks

    Another way of doing this is to differentiate the expression and set the first derivative to equal zero in order to find the value of x that minimises y. You can then see if this makes f(x) => 0.

    Let y = ln(u) where u = (3x^2 + 5x + 3). du/dx = 6x + 5, dy/du = 1/(3x^2 + 5x + 3).
    so dy/dx = (6x + 5) / (3x^2 + 5x + 3). dy/dx = 0 when 6x + 5 = 0, so 6x = -5, and x = -5/6.

    Sub in x = - 5/6 and you have your proof by counter-example.
 
 
 
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Updated: January 4, 2017
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