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hartnn Group TitleBest ResponseYou've already chosen the best response.1
\[f(x)=\sqrt{x^2+x+2}\sqrt{x^2+x+1}\]
 one year ago

Yahoo! Group TitleBest ResponseYou've already chosen the best response.0
Range = Real Numbers
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
I think dw:1347709678855:dw
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
what is ur approach?
 one year ago

estudier Group TitleBest ResponseYou've already chosen the best response.0
always positive, diff for max
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
x^2+x+2=(x+1/2)^2 +7/4 x^2+x+1=(x+1/2)^2+3/4 Now, f(x) is maximum when (x+1/2)^=0 because the f(x) value decresases as (x+1/2)^2 is greater than 0 since dw:1347710167952:dw Now, when (x+1/2)^2 >approaches infinity f(x) will be minimum but will never be zero.....but will be very close to zero
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Sorry not good at expalining things
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
But I think thats it
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
\[f(x)=\frac{1}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Is that another question?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
no this is same
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.1
\[f(x)=\sqrt{x^2+x+2}\sqrt{x^2+x+1}\times\frac{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.1
just rationalizing the denom, isn't it ?
 one year ago

sauravshakya Group TitleBest ResponseYou've already chosen the best response.3
Oh ya
 one year ago
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