Let f(x)=16/x+x-5=16x^{-1}+x-5f(x)=16x+x−5=16x−1+x−5. Then the derivative is f'(x)=-16x^{-2}+1=1-16/(x^2). Setting this equal to zero gives 16/(x^2)=1 so that x^2=16 and x=pm 4. These are the two critical points.
The second derivative is f''(x)=32x^{-3}=32/(x^3) so that f''(-4)<0 and f''(4)>0. This means there is a local maximum value at x=-4 and a local minimum value at x=4 (by the Second Derivative Test).
This also could have been derived from the First Derivative Test, in which we can also see that f'(x)\geq 0 when x\leq -4 and when x\geq 4 and f'(x)\leq 0 when -4\leq x<0 and when 0 < x\leq 4. This information also tells us that the function is increasing on the intervals (-infty,-4] and [4,infty) and is decreasing on the intervals [-4,0) and (0,4].
The local maximum value at x=-4 is f(-4)=-4-4-5=-13 and the local minimum value at x=4 is f(4)=4+4-5=3.
The graph of f is shown below.
graph{16/x+x-5 [-80, 80, -40, 40]}