Let #f(x)=16/x+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]}