# Monotonically increasing

• Nov 4th 2010, 02:28 AM
wilbursmith
Monotonically increasing
Here´s the question.
Attachment 19576

Ok so the derivative of f

${\frac {d}{dx}}f \left( x \right) =\cos \left( x \right) -1+3\,\alpha
\,{x}^{2}
$

The function is monotonically increasing if the derivative of f is bigger than zero in the interval. Or if $\alpha<1/3\,{\frac {1-\cos \left( x \right) }{{x}^{2}}}$

So my task should be to determine the lowest value of the right hand side of the inequality and say that alpha should be less than that? Is it so? I´ve tried that but ended up with zeros in the denominator and stuff. Any ideas?
• Nov 4th 2010, 03:30 AM
Also sprach Zarathustra
min(cosx-1) in [0,pi/2] = -1

max(cosx-1) in [0,pi/2] = 0
• Nov 4th 2010, 04:45 AM
wilbursmith
Quote:

Originally Posted by Also sprach Zarathustra
min(cosx-1) in [0,pi/2] = -1

max(cosx-1) in [0,pi/2] = 0

Yes. min(cosx-1) in [0,pi/2] =-1 and this happens as x --> pi/2 and therefore f´(0) larger than zero if

$4/3\,{\pi }^{-2}<\alpha$

But what happens when x-->0 and how about all the other values between 0 and pi/2?

I´m still not sure how to choose alpha so that f´(x) is larger than zero in the interval.