# Stationary points

• Nov 24th 2011, 03:04 PM
Furyan
Stationary points
Hello all

I have a couple of questions.

First, if the first derivative of a function is a quadratic can I show that the function has no stationary points by showing that the discriminant of the first derivative is less than zero?

Second, how do I go about finding the range of the values of a for which
y = x - (a/x) has no stationary points.

I know the answer is that a is greater than or equal to zero, but I don't know how to get there.

Thank you.
• Nov 24th 2011, 03:14 PM
pickslides
Re: Stationary points
Yep, that's correct, no stationary points within the reals.

For the second question, what did you get as the derivative?
• Nov 24th 2011, 03:19 PM
Furyan
Re: Stationary points
Thank you

I got 1+ ax^-2
• Nov 24th 2011, 03:31 PM
pickslides
Re: Stationary points
That is correct, what conditions can we put on 'a' for it to have zeros?

• Nov 24th 2011, 04:03 PM
Furyan
Re: Stationary points
Hello

I can see, I think, that if a = 0 then y=1 and that if a > 0, y > 1. When a < 0 I'm not sure.
• Nov 24th 2011, 04:17 PM
pickslides
Re: Stationary points
Quote:

Originally Posted by Furyan
When a < 0 I'm not sure.

In this case there will be zeros. pick a few values i.e a=-1, -3, -100 etc. It will become clear.
• Nov 25th 2011, 05:10 AM
Furyan
Re: Stationary points
Yes it has, crystal. I don't know why I didn't just do that in the first place. Thank you for your help.