k/32 (x2) 0<x>2 fx 3/32 (6-x) 2<x>6 0 otherwise
Find what K is?
Can anyone help with this question?
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?