Find the derivative using the chain rule. Let #y = sqrt(u)# and #u = 4 - x#. Then #(du)/(dx)= -1# and #dy/(du)= 1/(2sqrt(u))#

#dy/dx = -1 * 1/(2sqrtu)#

#dy/dx= -1/(2sqrt(4 - x))#

The trick now is to find critical numbers. These will occur when the derivative equals #0# or is undefined. The derivative has a horizontal asymptote at #y = 0#, so there will be no point on the derivative where it equals #0#. It is undefined however at #x = 4#.

Now let's test to see which side is increasing and which side is decreasing, and accordingly, whether #x= 4# is an absolute maximum or an absolute minimum.

**Test Point: #x = 3#**

#dy/dx = -1/(2sqrt(4 - 3)) = -1/(2(1)) = -1/2#

This means that in the interval #(-oo, 4)#,the function is decreasing and that at #x= 4#, there is an absolute minimum. This is also an endpoint, as the function has domain #{x| x ≤ 4, x in RR}#.

Therefore, the largest open interval #f(x)# is decreasing on is #(-oo, 4)#.

Hopefully this helps!