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anonymous
 4 years ago
Range of\[f(x)=\sqrt{x^2+x+2}\sqrt{x^2+x+1}\]
anonymous
 4 years ago
Range of\[f(x)=\sqrt{x^2+x+2}\sqrt{x^2+x+1}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I think dw:1347709678855:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0always positive, diff for max

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0x^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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry not good at expalining things

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\frac{1}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Is that another question?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\sqrt{x^2+x+2}\sqrt{x^2+x+1}\times\frac{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]

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