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# Decreasing functions question(Differentiation) watch

1. '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.
2. You can just say 6x^2+3>0 which is trivially true.
3. (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?
4. (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.
5. (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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