Find the range of the values for for which is strictly increasing.

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- October 19th 2010, 10:57 PMSambitproblem of increasing function
Find the range of the values for for which is strictly increasing.

- October 19th 2010, 11:43 PMGusbob
Find f'(x). For f(x) to be strictly increasing, f'(x) > 0 for all x.

- October 19th 2010, 11:48 PMSambit
yes i know that. but writing leads to:-

. after this how can i solve this for ? - October 19th 2010, 11:53 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? - October 19th 2010, 11:58 PMSambit
a quadratic equation does not have real roots when the discriminant is < zero. so, here, , ie, . am i correct?

- October 20th 2010, 12:07 AMGusbob
yes

- October 20th 2010, 12:12 AMSambit
but tell me one thing....here we are considering the non-existence of any real root of the quadratic equation because we have a

__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? - October 20th 2010, 12:13 AMGusbob
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.

- October 20th 2010, 12:15 AMSambit
ok. so the fact is the original equation, ie, can never be MONOTONICALLY DECREASING, but may be decreasing in some interval. is it?

ok. i am waiting for your reply - October 20th 2010, 12:19 AMGusbob
- October 20th 2010, 12:21 AMSambit
ok.. thanks a lot for spending so much time for my thread :D...

- October 20th 2010, 12:23 AMGusbob
Don't worry about it. Just another procrastination technique I have.