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14:08
Now as u′(0)>0, it is clear that the function will be convex throughout the interval [0,∞), thus u(x),u′(x),u"(x) will all be positive in the interval. Both statements follow.
@DevanshBhardwaj How did you claimed the convexity?
It would be possible only if u''(x)>0 for all x in R.
There seems some hollowness in your solution!
14:42
@AasthaChoudhary As $u(0)>0$
$u" (0)=u(0)>0$
As $u"(0)>0$ and $u'(0)>0$, thus, $\int_0^xu'(t)\,dt>0$ for all $t>0$. I hope this much seems clear.
@AasthaChoudhary Convexity in the interval is what we are concerned about, not throughout $R$. Suppose the function $x^3$. It is convex throughout $R^+$ even though it is concave in $R^-$. So not necessary throughout R.
@AasthaChoudhary Sorry :(
15:05
Seems unclear @DevanshBhardwaj
Since this is only for a specific x=0.
Ok. So what should be the condition for $u(x)<0$?
The quest is for [0,inf)
for some x in the interval $[0,\infty)$
I am not getting you right?
Sorry for?
most probably, but that is because I am bad at explaining:)
@AasthaChoudhary Hollowness seems to mean a useless post
15:08
oho!
Ok so what should be the condition for $u(x)<0$
@DevanshBhardwaj no I meant things were unclear, not useless :)
For u'(x)<0 we can claim u(x) is strictly decreasing.
As $u"(0)>0$, thus the derivative at 0 is increasing right?
15:11
for u(x)<0 i.e. negative!
@AasthaChoudhary Well lets leave that. I found a better way to explain
ok so derivative is increasing
yes but this is for x=0 only :/
Well I think I should make it a little formal, which I am not able to think how right now. Indeed I understand why you are not able to understand me. Bye
Ill post it on this chatroom and in the answer

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