Find the range of the values for $\displaystyle c$ for which $\displaystyle f(x) = x^3 - 3x^2 + 3cx + 1$ is strictly increasing.

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- Oct 19th 2010, 09:57 PMSambitproblem of increasing function
Find the range of the values for $\displaystyle c$ for which $\displaystyle f(x) = x^3 - 3x^2 + 3cx + 1$ is strictly increasing.

- Oct 19th 2010, 10:43 PMGusbob
Find f'(x). For f(x) to be strictly increasing, f'(x) > 0 for all x.

- Oct 19th 2010, 10:48 PMSambit
yes i know that. but writing $\displaystyle f'(x) > 0$ leads to:-

$\displaystyle 3x^2 - 6x + 3c > 0$. after this how can i solve this for $\displaystyle c $ ? - Oct 19th 2010, 10:53 PMGusbob
Use the quadratic formula, pay attention in particular to the discriminant.

Think about what it means to have f'(x) > 0. It means if you graph f'(x), you should have every single point above the x axis => no real roots.

So under what conditions does a quadratic equation not have real roots? - Oct 19th 2010, 10:58 PMSambit
a quadratic equation does not have real roots when the discriminant is < zero. so, here, $\displaystyle 4 < 4c$, ie, $\displaystyle c>1$. am i correct?

- Oct 19th 2010, 11:07 PMGusbob
yes

- Oct 19th 2010, 11:12 PMSambit
but tell me one thing....here we are considering the non-existence of any real root of the quadratic equation because we have a

*$\displaystyle > 0$*condition. but in case we had the function__decreasing__, then also we would have considered the same criterion, that is, the non-existence of any real root; and hence the final answer (ie, c > 1) would be the same.

how can you explain this? - Oct 19th 2010, 11:13 PMGusbob
If we had a decreasing function (to be more precise, polynomials), you cannot have a positive leading coefficient for for f'(x). Simply because the limit as x gets very large will be positive infinity.

- Oct 19th 2010, 11:15 PMSambit
ok. so the fact is the original equation, ie, $\displaystyle f(x) = x^3 - 3x^2 + 3cx + 1 $ can never be MONOTONICALLY DECREASING, but may be decreasing in some interval. is it?

ok. i am waiting for your reply - Oct 19th 2010, 11:19 PMGusbob
- Oct 19th 2010, 11:21 PMSambit
ok.. thanks a lot for spending so much time for my thread :D...

- Oct 19th 2010, 11:23 PMGusbob
Don't worry about it. Just another procrastination technique I have.