dan18 3 years ago The maximum value of the function

1. hartnn

can be found by taking the derivative of that funtion and equating it to 0.

2. dan18

y=$\sqrt{2-x}$

3. hartnn

so did u find y' ?

4. dan18

y=-4[\sqrt{2-x}\]

5. hartnn

could u differentiate y ?

6. miteshchvm

differentiate the function, then put the differentiated equation = 0 you will have either one answer or 2, if you get two answers, the one which has negative value will be the maximum value of the function ;)

7. dan18

is it the y value?

8. hartnn

have u learned to differentiate x^n ?

9. dan18

y'= nx^n-1

10. hartnn

so couldn't u differentiate y=-4 sqrt(2-x) ??

11. dan18

does the max. value asked here is the x or the y value?

12. hartnn

y value

13. dan18

y'= 2/(2-x)^-(1/2)

14. akash_809

@dan18 , the correct way is to differentiate and equate to zero , but just by looking at this question once can tell it's maximum value is infinity for x=-infinity....pretty weird question i guess...are u sure u hv typed the question correctly

15. hartnn

wouldn't it be y'= 2/((2-x)^(1/2))

16. dan18

I get x=2 @ y', then subts. it to the eq. does the max. value = 0?

17. akash_809

@dan18 is the question correct ?

18. hartnn

but actual answer is infact Max. value = 0 when x=2 http://www.wolframalpha.com/input/?i=y%3D-4+sqrt%282-x%29

19. dan18

$y=-4\sqrt{2-x}$

20. dan18

thak you guys!

21. akash_809

now u hv got the correct question, yes the max would be zero