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phidya
Student XII IPA
w 12 Fungsi f(x) = dengan selang 0 <x< 21 mencapai nilai minimum a 1-2cos2x pada beberapa titik X1. Nilai a + adalah A. 10 B. 15 C. 16 D. 18 E. 20
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Putra Gema Nusa
Rockstar Teacher GURU SMA
m # # nilai minimum a=-12 do x, = {0°, 180°, 360°3 = {0, 1,2173 at AXI # Cari nilai 4x ㅠ » X,=0 -12 + 4.0 T 6-12 atau -8 - -12 to = - -12 atau -a n. :> X, - - -12 t 4.T TT = -127 4 = -8 > Xi = 2T -0 -12 t 4.27 ㅠ = -1278 lu -4 Pahamify 
12 h 1-2cos 2x x - k. 180° f(x) = O EX < 2T minimum a di beberapa titik X, 4X, ㅠ # Syarat minimum/maksimum f'(X) = 0 9 Ditanya at » 2X = 0 + k. 360V -> 2x = 180'+k. 360° °tk х k x = go'tk. 180° k=0 ox=0 k=0x=90° k= 1 -> x = 180° k=1 X=270 k= 2 + x = 360° # x = {0,900, 180°, 270°, 360°3 12 -12 1-2 los o |-2605 720 12 4 =-12 1 - 2005 180 f(x) = 12 (1–2 65 2x)" •> f(0) =_12 E > of(366 ) = - f'(X) = – 12 (1– 2482x)"?. (0–2.26- 4n zx) ( sin U os =f(90) = こ こ 2 > f(180) =_12 -12 -12.4 sin 2x o 0 (1–2 605 2x) -48 sin 2x 0 (1-2003 2x) * ) sin 2x=0 - Sin 2x = sino 1-203 360° 2 12 ?f(270) = 4 1-265 540° Pahamify
m # # nilai minimum a=-12 do x, = {0°, 180°, 360°3 = {0, 1,2173 at AXI # Cari nilai 4x ㅠ » X,=0 -12 + 4.0 T 6-12 atau -8 - -12 to = - -12 atau -a n. :> X, - - -12 t 4.T TT = -127 4 = -8 > Xi = 2T -0 -12 t 4.27 ㅠ = -1278 lu -4 Pahamify 
12 h 1-2cos 2x x - k. 180° f(x) = O EX < 2T minimum a di beberapa titik X, 4X, ㅠ # Syarat minimum/maksimum f'(X) = 0 9 Ditanya at » 2X = 0 + k. 360V -> 2x = 180'+k. 360° °tk х k x = go'tk. 180° k=0 ox=0 k=0x=90° k= 1 -> x = 180° k=1 X=270 k= 2 + x = 360° # x = {0,900, 180°, 270°, 360°3 12 -12 1-2 los o |-2605 720 12 4 =-12 1 - 2005 180 f(x) = 12 (1–2 65 2x)" •> f(0) =_12 E > of(366 ) = - f'(X) = – 12 (1– 2482x)"?. (0–2.26- 4n zx) ( sin U os =f(90) = こ こ 2 > f(180) =_12 -12 -12.4 sin 2x o 0 (1–2 605 2x) -48 sin 2x 0 (1-2003 2x) * ) sin 2x=0 - Sin 2x = sino 1-203 360° 2 12 ?f(270) = 4 1-265 540° Pahamify
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