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2 Diketahui f(x) = sin (axtboº I Tentukan absis titik stasioner b. interval fungsi naik ។ untuk olx L 360° 3 Tentukan interval fungsi turun pada fungsi y= sinx & cosx untuk ocx < 360 A Tentukan koordinat titik stasioner pada fungsi f(x) = sin²x untuk OLX < 360°
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3) Y - Çin x + os x , 0 < x < 360° y= x tas X ľ X < Ditanya: interval fungs y'= X - x # Syarat fungsi turun y'<i turun X uns-X SM = 0 > XUIS – XSCO a a los X - sinx Sinx = Wsx + + 0.0 45° eszz 360° | = s Sinx cosx fan x = tan 450 x = 450 tk. 180° >k=0 X=45° > k=1 X = 225' & interval y turun 45° < x < 225° o . Pahamify 
A 4 f(x) = sin X, US = (st on x # ,5.0 1 a o= Xc flol=0 1 = =0 - ²x 0 < x < 360 # Subtitusi nilai-illal x Ditanya koordinat titik stasioner =... :> X = go of f (90°) = 1 Syarat! f'(X) = 0 ! (x > X = 180° -$180) 0 f'() - 2 sinx.asx f(360° ) {x,y)(0,0),600,1),(180,0), (270".1). (360%,0) () 1= (otzita etz axc. :> X = 360° 0 = 0 = XZ uis 0 XZ U sin 2x = sin o o # 2X=0 tk.360° # 24 = 180°tk. 360° X = k. 180° X = 90° tk. 180° sk=0 X=0 >k=00x=90° >kal » X= 180° sk=10X=270° 7 k= 2 X=360° o Pahamify 
2x = -150 tl.360° :: Oʻ< x <15 atau 105'<x< 1950 atau a 2 flx) = sin (2X +60), 0 €* 360* a. absis fotik stasioner (f'(x) =0) + 2x + 60°= – 900+ K.360° f'(x) = 2 6s (2X+60) x = -75°tk. 180° 2 cos (2x+60°) = 0 os (2x + 60°) = 0 cos (2x +60°) -. X = {is", 105°, 1959, 285"} # 2x+60° = 90' + k.360° b. Interval fungri nake ( f(x) >0) f'Q - = 2 aas (2x + 60') >0 -> k = 1 - X = 105 "k=208= 285° 285 LX < 360° = cos go Х s s () 2X = 30° tk. 360° x = 19 tk. 180 -> K+0 -0 =15 > K=10x = 1959 Đ + + + 0° I, げ 105 195 28 Dahamify 360
3) Y - Çin x + os x , 0 < x < 360° y= x tas X ľ X < Ditanya: interval fungs y'= X - x # Syarat fungsi turun y'<i turun X uns-X SM = 0 > XUIS – XSCO a a los X - sinx Sinx = Wsx + + 0.0 45° eszz 360° | = s Sinx cosx fan x = tan 450 x = 450 tk. 180° >k=0 X=45° > k=1 X = 225' & interval y turun 45° < x < 225° o . Pahamify 
A 4 f(x) = sin X, US = (st on x # ,5.0 1 a o= Xc flol=0 1 = =0 - ²x 0 < x < 360 # Subtitusi nilai-illal x Ditanya koordinat titik stasioner =... :> X = go of f (90°) = 1 Syarat! f'(X) = 0 ! (x > X = 180° -$180) 0 f'() - 2 sinx.asx f(360° ) {x,y)(0,0),600,1),(180,0), (270".1). (360%,0) () 1= (otzita etz axc. :> X = 360° 0 = 0 = XZ uis 0 XZ U sin 2x = sin o o # 2X=0 tk.360° # 24 = 180°tk. 360° X = k. 180° X = 90° tk. 180° sk=0 X=0 >k=00x=90° >kal » X= 180° sk=10X=270° 7 k= 2 X=360° o Pahamify 
2x = -150 tl.360° :: Oʻ< x <15 atau 105'<x< 1950 atau a 2 flx) = sin (2X +60), 0 €* 360* a. absis fotik stasioner (f'(x) =0) + 2x + 60°= – 900+ K.360° f'(x) = 2 6s (2X+60) x = -75°tk. 180° 2 cos (2x+60°) = 0 os (2x + 60°) = 0 cos (2x +60°) -. X = {is", 105°, 1959, 285"} # 2x+60° = 90' + k.360° b. Interval fungri nake ( f(x) >0) f'Q - = 2 aas (2x + 60') >0 -> k = 1 - X = 105 "k=208= 285° 285 LX < 360° = cos go Х s s () 2X = 30° tk. 360° x = 19 tk. 180 -> K+0 -0 =15 > K=10x = 1959 Đ + + + 0° I, げ 105 195 28 Dahamify 360
3) Y - Çin x + os x , 0 < x < 360° y= x tas X ľ X < Ditanya: interval fungs y'= X - x # Syarat fungsi turun y'<i turun X uns-X SM = 0 > XUIS – XSCO a a los X - sinx Sinx = Wsx + + 0.0 45° eszz 360° | = s Sinx cosx fan x = tan 450 x = 450 tk. 180° >k=0 X=45° > k=1 X = 225' & interval y turun 45° < x < 225° o . Pahamify 
A 4 f(x) = sin X, US = (st on x # ,5.0 1 a o= Xc flol=0 1 = =0 - ²x 0 < x < 360 # Subtitusi nilai-illal x Ditanya koordinat titik stasioner =... :> X = go of f (90°) = 1 Syarat! f'(X) = 0 ! (x > X = 180° -$180) 0 f'() - 2 sinx.asx f(360° ) {x,y)(0,0),600,1),(180,0), (270".1). (360%,0) () 1= (otzita etz axc. :> X = 360° 0 = 0 = XZ uis 0 XZ U sin 2x = sin o o # 2X=0 tk.360° # 24 = 180°tk. 360° X = k. 180° X = 90° tk. 180° sk=0 X=0 >k=00x=90° >kal » X= 180° sk=10X=270° 7 k= 2 X=360° o Pahamify 
2x = -150 tl.360° :: Oʻ< x <15 atau 105'<x< 1950 atau a 2 flx) = sin (2X +60), 0 €* 360* a. absis fotik stasioner (f'(x) =0) + 2x + 60°= – 900+ K.360° f'(x) = 2 6s (2X+60) x = -75°tk. 180° 2 cos (2x+60°) = 0 os (2x + 60°) = 0 cos (2x +60°) -. X = {is", 105°, 1959, 285"} # 2x+60° = 90' + k.360° b. Interval fungri nake ( f(x) >0) f'Q - = 2 aas (2x + 60') >0 -> k = 1 - X = 105 "k=208= 285° 285 LX < 360° = cos go Х s s () 2X = 30° tk. 360° x = 19 tk. 180 -> K+0 -0 =15 > K=10x = 1959 Đ + + + 0° I, げ 105 195 28 Dahamify 360
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