Calculus of a Single Variable, Chapter 3, 3.4, Section 3.4, Problem 24

Given the function f(x)=(x+3)/sqrt(x)
i.e it can be written as f(x)=x/sqrt(x)+3/sqrt(x) =x^(1/2)+3x^(-1/2)
Taking the first derivative we get,
f'(x)=1/2x^(-1/2)-3/2x^(-3/2)
Now again differentiating we get,
f''(x)=-1/4x^(-3/2)+9/4x^(-5/2)
Now the points of inflection can be found out by equating f''(x) to zero. i.e.
f''(x)=0
i.e. -1/4x^(-3/2)+9/4x^(-5/2)=0
9x^(-5/2)-x^(-3/2)=0
x^(-3/2)(9/x-1)=0
x^(-3/2)((9-x)/x)=0
(9-x)/x^(5/2)=0
Therefore x=9 is the inflection point.

Now choose auxilliary points x=8 to the left of inflection point and x=10 to the right of inflection point.
Sof''(8)=0.0013>0
Therefore on (-infinity, 9) , the curve is concave upward .
Now,
f''(10)=-0.0007<0
Therefore on (9, infinity) , the curve is concave downward.