Question

Find the absolute extrema of the function on the closed interval. h(s) = 10 / (s − 3) , [0, 1]

Answer 1

I find the absolute hs) = 1 extrema of , [0, 1] the critical point first find his) 0 (902)) = 9(5): U= 5-3, flu) = Stw.9159 dut (5-3) =10 - 0 (u'). du u=5-3 -10 (5-32 h13)=0 (5-3) = 0 5-3=0 s=3 – Critical point Critical point is not in the given closed so, We have to check with or Interval

at s=0 h(s) = 1 (0) = 10 = -10 = -3.33 at s=1 1-3 111 Absolute extrema at [3= o -5 h(s)= (-3.33]