k/32 (x ^{2)}0<x>2 fx 3/32 (6-x) 2<x>6 0 otherwise

Find what K is?

Can anyone help with this question?

Results 1 to 5 of 5

- Apr 30th 2013, 12:41 PM #1

- Joined
- Aug 2012
- From
- uk
- Posts
- 24

- Apr 30th 2013, 12:47 PM #2

- Apr 30th 2013, 12:57 PM #3

- Joined
- Aug 2012
- From
- uk
- Posts
- 24

- Apr 30th 2013, 01:12 PM #4

- Joined
- Apr 2005
- Posts
- 18,957
- Thanks
- 2740

## Re: Can anyone help?

If you are serious then you have real problems. There is no reason why "k" cannot have any value at all or even "no" value.

Now if the problem were to "find the value of k such that the function defined is continuous" then there would be one unique value for k.

Does the problem say anything about "cotinuous"? Do you know what "continuous" means?

- Apr 30th 2013, 01:27 PM #5

- Joined
- Aug 2012
- From
- uk
- Posts
- 24