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Find the absolute extrema. 3 3 2 (a) f(x) = x , - - 2 x (b) f(t) = 32/32t, [1,2] [-1,1] 6x2 (c)
Find the absolute extrema. 3 3 2 (a) f(x) = x , - - 2 x (b) f(t) = 32/32t, [1,2] [-1,1] 6x2 (c) g(x): = [-2,1] X 2' (d) (s) = |2s 1],[0,2] - (e) f(0) = 3 cos 0, [0,2] (782), [0,2] (f) g(x) = tan (g) f(x) = 2 cos x - cos 2x, [0,2]
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Thomas Calculus Early Transcendentals
Authors: Joel R Hass, Christopher E Heil, Maurice D Weir
13th Edition
978-0321884077, 0321884078
