Question

Locate the absolute extrema of the function on the closed interval:

y= 4/x+tan(pix/8), [1,2]

Answer 1

dy/dx = -4/x^2 + sec^2 (πx/8) (π/8)
= 0 for a max/min

π/(8cos^2 (πx/8)) = 4/x^2

8cos^2 (πx/8) /π = x^2/4

cos^2 (πx/8) = πx^2/32

there is no simple way to solve,
I ran it through Wolfram's amazing equation solver and got

x = ± 2.13448

http://www.wolframalpha.com/input/?i=cos%5E2+%28πx%2F8%29+%3D+πx%5E2%2F32

arrghhh, after all that, our solution is ouside the domain, so simply

if x=1
y = 4/1 + tan(π/8) = 4.414
if x = 2
y = 4/2 + tan(π/4) = 2+1 = 3

look at the graph of

http://www.wolframalpha.com/input/?i=y+%3D+4%2Fx+%2B+tan%28πx%2F8%29

everything makes sense.