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    'Show that f(x) = 4-x(2x^2+3) is decreasing for all real numbers.

    By differentiating and forming the inequality you get -6x^2-3<0.

    Since you can't solve this, doesn't this mean the function is never decreasing, since the gradient cannot be less than 0 unless you don't use real numbers.
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    You can just say 6x^2+3>0 which is trivially true.
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    (Original post by thekidwhogames)
    You can just say 6x^2+3>0 which is trivially true.
    Good point! Thanks. But wouldn't that mean for all real numbers, the gradient would be >0, and so wouldn't always be decreasing?
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    (Original post by dont know it)
    Good point! Thanks. But wouldn't that mean for all real numbers, the gradient would be >0, and so wouldn't always be decreasing?
    'Show that f(x) = 4-x(2x^2+3) is decreasing for all real numbers.

    f(x) = 4-2x^3 - 3x --> f'(x) = -6x^2 - 3.

    We know 6x^2 + 3 is always positive meaning f'(x), which is that multiplied by -1, will always be negative so always decreasing.

    Hope this helped.
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    (Original post by thekidwhogames)
    'Show that f(x) = 4-x(2x^2+3) is decreasing for all real numbers.

    f(x) = 4-2x^3 - 3x --> f'(x) = -6x^2 - 3.

    We know 6x^2 + 3 is always positive meaning f'(x), which is that multiplied by -1, will always be negative so always decreasing.

    Hope this helped.
    Oh yes, thank you.
 
 
 
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