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mukushla

  • 3 years ago

Range of\[f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\]

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  1. hartnn
    • 3 years ago
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    \[f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\]

  2. mukushla
    • 3 years ago
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    yeah

  3. Yahoo!
    • 3 years ago
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    Range = Real Numbers

  4. sauravshakya
    • 3 years ago
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    I think |dw:1347709678855:dw|

  5. mukushla
    • 3 years ago
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    thats it

  6. mukushla
    • 3 years ago
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    what is ur approach?

  7. estudier
    • 3 years ago
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    always positive, diff for max

  8. sauravshakya
    • 3 years ago
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    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

  9. sauravshakya
    • 3 years ago
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    Sorry not good at expalining things

  10. sauravshakya
    • 3 years ago
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    But I think thats it

  11. mukushla
    • 3 years ago
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    \[f(x)=\frac{1}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]

  12. sauravshakya
    • 3 years ago
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    Is that another question?

  13. mukushla
    • 3 years ago
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    no this is same

  14. sauravshakya
    • 3 years ago
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    ?

  15. mukushla
    • 3 years ago
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    \[f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\times\frac{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}\]

  16. hartnn
    • 3 years ago
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    just rationalizing the denom, isn't it ?

  17. mukushla
    • 3 years ago
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    yes

  18. sauravshakya
    • 3 years ago
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    Oh ya

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