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

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

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

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

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

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

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

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

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

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

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

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