Question

Find the absolute extrema of the function on the closed interval. (Order your answers from smallest to largest x, then from smallest to largest y.) f(x) = sin(2x), (0, 2) minimum (x,y) = (x,y) - maximum (x, ) = (x,y) =

Answer 1

# f(x) = sin(2x), (0, 2TT). then f'(x) = f(x) = o Sin(2x) = COS(2x)..(2x) = cos(2x) (2011) = 2 Cos (21) dx dx possible critical we = f'(x) = 2 Cos(2x) numbers of f, we equste f'cas to o, To find the get, > COS(2x)=0 = cos(2x) = 0 => Cos(20) = cos((2*+1)) in0,1,2,--- x = n=0,1,2,-.. (20+1) (2241 ) I LA i mall, --- X a 3, Sm, 7, 9, But the only values of a in Co, 25] Y are ST 2 ना x-values value of f(x) = sin(2x) Remarte 77 4 values Absolute mana 37 4 Absolute min. 571 Absolute m4n. 7 -1 Absolute mini 0 END points of Co, 2] 27 0

minionem (३, ४) (34 -1) प, -1) (४,५) %3D maximum (x,8) (1) (४५) 3D [(म प) is the desired answer
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