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$y=x^2$ and $D_f=(0,2]$.

So, my interval is $0<x^2<=2$

But, according to my book it has no minimum values. But, it has maximum value which is $2^2=4$

According to me, it should have minimum value either. Cause, $1$ exists in that interval also.So, $1$ is minimum value. But, why they said that there's no minimum value?

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    What does it have to do with $1$?? $0.5$ is also there and what? – Ѕᴀᴀᴅ Jul 14 '21 at 06:38
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    @Saad $1$ is minimum value. That's what I think. My question was why my book said there's no minimum value? –  Jul 14 '21 at 06:38
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    We have $f(D_f)=(0,4]$. Hence $ \inf f(D_f)$ exists and $=0.$ But $ \min f(D_f)$ does not exist. – Fred Jul 14 '21 at 06:40
  • @logloglogx 's comment was useful for me. But, he had removed that –  Jul 14 '21 at 06:42

1 Answers1

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This function is increasing on this interval (prove it).

Suppose the minimum occured at $x > 0$. Consider $x/2$ - uh-oh.

See the problem? No minimum exists on $(0, 2].$

Sean Roberson
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