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.