differentiating question: let f(x)=x^3-3x^2+kx+8 where k is a constant.

A

AlonzoN

Junior Member
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Jul 31, 2021
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I need help understanding this. I got most of the work done:

I differentiated f(x), and so I got f'(x) = 3x^2 - 6x + k

And so I know that f'(x) > 0 for it to be an increasing function.

(Mind you, this question is more than 5 years old, there was another post with the exact same thing, but the working out was all jumbled, and I couldn't understand it).

I just REALLY need help with this.
 
You have not really asked a question, at least not a clear one. Do you want f'(x) to ALWAYS be greater than 0? In the post that was 5 years old I recall that staple was saying that f'(x) was only positive (actually I think negative) for some of the x-values.
 
For a cubic to always be increasing, that is no max or min values, you need b^2-3ac = 9 - 3k < 0. So k > 3
 
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