Thread: Third Derivitive of Decreasing Function

1. Third Derivitive of Decreasing Function

Hi guys, im stuck on a problem, this is it:
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Let f(x) = - u''(x)/u'(x)
Show that if f(x) is decreasing then u'''(x) is greater than 0.

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Thanks Guys, ive been at it for a while and cant get it

2. Originally Posted by furnis1
Hi guys, im stuck on a problem, this is it:
------

Let f(x) = - u''(x)/u'(x)
Show that if f(x) is decreasing then u'''(x) is greater than 0.

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Thanks Guys, ive been at it for a while and cant get it
If $f = - \frac{u''}{u'}$ then $f' = - \left( \frac{u' u''' - u''^2}{u'^2} \right)$. If $f' < 0$ then $- \left( \frac{u' u''' - u''^2}{u'^2} \right) < 0$ so $\left( \frac{u' u''' - u''^2}{u'^2} \right) > 0$ so $u' u''' - u''^2 > 0$ giving $u' u''' > u''^2$ so $u' u''' > 0$. You need the product. It is possible to come up with a function that $u' > 0\;\;\text{and}\;\;\; u'''<0$.

Case in point $u ' = e^{-x^2}$ so $f = -2x\;\;\Rightarrow\;\; f' = -2 < 0$ yet $u''' = (4x^2-2)e^{-x^2}$ which is not always positive for all x