## mukushla 3 years ago Range of$f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}$

1. hartnn

$f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}$

2. mukushla

yeah

3. Yahoo!

Range = Real Numbers

4. sauravshakya

I think |dw:1347709678855:dw|

5. mukushla

thats it

6. mukushla

what is ur approach?

7. estudier

always positive, diff for max

8. sauravshakya

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

Sorry not good at expalining things

10. sauravshakya

But I think thats it

11. mukushla

$f(x)=\frac{1}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}$

12. sauravshakya

Is that another question?

13. mukushla

no this is same

14. sauravshakya

?

15. mukushla

$f(x)=\sqrt{x^2+x+2}-\sqrt{x^2+x+1}\times\frac{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}{\sqrt{x^2+x+2}+\sqrt{x^2+x+1}}$

16. hartnn

just rationalizing the denom, isn't it ?

17. mukushla

yes

18. sauravshakya

Oh ya