+14985 Hello Guest!
Where is f(x) = (1/32)sqrt(x) + 1/x, x>0, concave up and concave down?
\({\color{blue}f(x)=\frac{1}{32}\times\sqrt{x}+\frac{1}{x} }\) x > 0
The graph is downwards convex in the range x> 0.
Greeting asinus :- ) 
!
Sorry Asinus, but that graph is misleading and your conclusion is incorrect.
For small values of x the function behaves as 1/x and the graph will be much as shown.
For large values of x though the 1/x term becomes insignificant and the function behaves as the
sqrt (x)/32 term and this will tend to infinity as x tends to infinity.
It follows that the function must have a minimum at some point.
To find the minimum, put the first derivative of f (x) equal to zero, (it's at x = 16).
Further, the curve is concave up to begin with but then switches to concave down.
To find the point at which the concavity changes, put the second derivative equal to zero, (it occurs at approximately x = 40.3).
