anonymous
  • anonymous
Range of\[f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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hartnn
  • hartnn
\[f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\]
anonymous
  • anonymous
yeah
anonymous
  • anonymous
Range = Real Numbers

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anonymous
  • anonymous
I think |dw:1347709678855:dw|
anonymous
  • anonymous
thats it
anonymous
  • anonymous
what is ur approach?
anonymous
  • anonymous
always positive, diff for max
anonymous
  • anonymous
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
anonymous
  • anonymous
Sorry not good at expalining things
anonymous
  • anonymous
But I think thats it
anonymous
  • anonymous
\[f(x)=\frac{1}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]
anonymous
  • anonymous
Is that another question?
anonymous
  • anonymous
no this is same
anonymous
  • anonymous
?
anonymous
  • anonymous
\[f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\times\frac{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]
hartnn
  • hartnn
just rationalizing the denom, isn't it ?
anonymous
  • anonymous
yes
anonymous
  • anonymous
Oh ya

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