Find the first derivative of the function.
If #f'(5)<0#, then #f(x)# is decreasing when #x=5#.
If #f'(5)>0#, then #f(x)# is increasing when #x=5#.
To find #f'(x)#, use the quotient rule.
#f'(x)=(sqrt(x+3)d/dx[x]-xd/dx[sqrt(x+3)])/(sqrt(x+3))^2#
Find each derivative separately.
#d/dx[x]=1#
The following derivative requires the chain rule.
#d/dx[sqrt(x+3)]=1/2(x+3)^(-1/2)overbrace(d/dx[x+3])^(=1)=1/(2sqrt(x+3))#
Plug the derivatives back in.
#f'(x)=(sqrt(x+3)-x/(2sqrt(x+3)))/(x+3)#
Multiply everything by #2sqrt(x+3)# to clear out the denominator of the internal fraction.
#f'(x)=(2(x+3)-x)/(2(x+3)sqrt(x+3))#
#f'(x)=(x+6)/(2(x+3)^(3/2))#
Now, find #f'(5)#.
#f'(5)=(5+6)/(2(5+3)^(3/2))=11/(2(8)^(3/2))#
We could determine the exact value of #f'(5)#, but it should be clear that the answer will be positive.
Thus, #f'(x)# is increasing when #x=5#.
