Thread: problem of increasing function

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

Find f'(x). For f(x) to be strictly increasing, f'(x) > 0 for all x.

yes i know that. but writing leads to:-
. after this how can i solve this for ?

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?

a quadratic equation does not have real roots when the discriminant is < zero. so, here, , ie, . am i correct?

yes

but tell me one thing....here we are considering the non-existence of any real root of the quadratic equation because we have a 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?

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.

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

Originally Posted by Sambit

ok. so the fact is the original equation, ie, can never be MONOTONICALLY DECREASING, but may be decreasing in some interval. is it?

Yes it will decrease at some interval, but never monotonic decreasing.

ok.. thanks a lot for spending so much time for my thread ...

Don't worry about it. Just another procrastination technique I have.