Finde das absolute Extrem der gegebenen Funktion #f (x) = sinx + cosx # im Intervall # [0,2pi] #?

Antworten:

# f_max = sqrt2 #
# f_min = -sqrt2 #

Erläuterung:

#f '(x) = cosx-sinx #

#f '(x) = 0 <=> cosx-sinx = 0 #

# cosx / cosx-sinx / cosx = 0 ^^ cosx! = 0 #

# tanx = 1 ^ ^ cosx! = 0 #

# x = pi / 4 + kpi ^^ x! = pi / 2 + mpi #

# x = pi / 4 + kpi #

Auf # [0,2pi] # :

# x = pi / 4 vv x = (5pi) / 4 #

# f_max = f (pi / 4) = sin (pi / 4) + cos (pi / 4) = sqrt2 / 2 + sqrt2 / 2 = sqrt2 #

# f_min = f ((5pi) / 4) = sin ((5pi) / 4) + cos ((5pi) / 4) = -sqrt2 / 2-sqrt2 / 2 = -sqrt2 #

# f_max = sqrt2 #
# f_min = -sqrt2 #