We start by differentiating.
Let #y = u^(1/2)# and #u = x + 3#. By the power rule, #y' = 1/2u^(-1/2) = 1/(2u^(1/2))# and #u' = 1#
By the chain rule:
#dy/dx= dy/(du) xx (du)/dx#
#dy/dx = 1/(2u^(1/2)) xx 1#
#dy/dx = 1/(2(x + 3)^(1/2))#
For #f'(x)#, we find the derivative by the quotient rule:
#f'(x) = (1sqrt(x+ 3) - (x - 3)1/(2(x + 3)^(1/2)))/(sqrt(x+ 3))^2#
#f'(x) = (sqrt(x + 3) - (x - 3)/(2(x + 3)^(1/2)))/(x + 3)#
#f'(x) = ((2sqrt(x + 3)sqrt(x+ 3) - x + 3)/(2(x + 3)^(1/2)))/(x+ 3)#
#f'(x) = ((2(x + 3) - x + 3)/(2(x+ 3)^(1/2)))/(x + 3)#
#f'(x) = ((2x + 6 - x + 3)/(2(x + 3)^(1/2)))/(x + 3)#
#f'(x) = (x + 9)/(2(x + 3)^(1/2)x + 3)#
#f'(x) = (x + 9)/(2(x + 3)^(3/2))#
Now, if the function is increasing at the point #x = a#, then #f'(a) > 0#. If the function is decreasing at #x= a#, then #f'(a) < 0#.
#f'(5) = (5 + 9)/(2(5 + 3)^(3/2))#
#f'(5) = 14/(2sqrt(512))#
#f'(5) = 14/(2(16)sqrt(2))#
#f'(5) = 14/(32sqrt(2)#
#f'(5) = 7/(16sqrt(2))#
What matters is that the above value is positive. Hence, #f(x)# is increasing at #x = 5#.
Hopefully this helps!
