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Didapat:
Maka, uji nilai masing-masing:
f(37) = 3sin(37) + 4cos(37)
= 3.(3/5) + 4.(4/5)
= 9/5 + 16/5
= 25/5
= 5
f(217) = 3sin(217) + 4cos(217)
= 3(-3/5) + 4(-4/5)
= -9/5 - 16/5
= -25/5
= -5
Maka, nilai minimumnya adalah -5
Nomor 2.
Dengan:
Didapat:
Akan maks/min ketika f'(x) = 0
-2cos(4x) = 0
cos(4x) = cos(90)
Solusi:
4x = 90
Dan,
4x = -90+360
4x = 270
Sehingga, uji masing-masing:
f(90) = 5 - ½sin(90)
= 5 - 1
= 4
f(270) = 5 - ½sin(270)
= 5 - (-1)
= 6
Maka, nilai minimum = 4