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) 2 2 Diketahui f(x) = sin (2x+60°) Tentukan : a. absis titik stasioner b. interval fungsi naik ។ untuk olx L 360° 3 Tentukan interval Fungsi turun pada fungsi y= sinx & cos x untuk BEX € 360° A Tentukan koordinat titik stasioner pada fungsi f(x) = sin²x untuk os x < 360°
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Putra Gema Nusa
Rockstar Teacher GURU SMA
3) y= sin x + OS X, Ò EX €360° Ditanya: interval fungu turun y': = Los X-SM X # Syarat fungsi turun y y'<i » WsX - SIN X <o 0 = los X - sinx + t Sinx = Wsx Sinx D 45 225 360° cosx 5 & interval y turun 45° <x<225° 2 fan x=tan 450 X = 450 tk. 180° ok=0 - X=45° > k=1 X = 225' = Pahamify 
A f(x) X₂ UiS -X sinex a 0 = Xc X = 1 0 X° D. 50 Oʻ = x < 360° # Subtitusi nilai - Millal x ke f(x) = an’x < Ditanya koordinat titik stasioner flo)=0 > X = 90' of (90°) = Syarat! f(x)=0 -> x = 180° -> $(180) 0 f'(x) - 2 sin x.asx f(360) ={{x,y)(0,0),(507,1) , (10%,0), (270".), # X=0tk.360° # 2X= 180°tk. 360° (360º,0)) 1 = (otzita ,otz exc. a 2 :> X > 360° - = 0 = Sin 2x 0 sin 2x = sin o 5 .081.77.06- X ,lb=XG 0=716 x - k. 180° sk=0 +x=0 ?K= 1 + x = 180° 7 k= 2 X=360° o sk=10x=270° Pahamify 
+ lo 0<x<15 atau 105°<x< 1959 atau . a ) 3 f(x) = sin (2X+60), 0 < x < 360° a. absis fitik stasioner (f '(x)=0) + 2x + 60°= -90°+k. 360° a. °= 2x = -150° +1.360° f'(x) - 2 cos (2X+60) X = -75°4k, 180° 2 cos (2x+60) = 0 -> k = 1 x = 105 as (2x + 60°) = 0 »k=20 X = 285° cos (2x+60) = cos go 3. X= { IS', 105°, 1959, 285"} # 2x+60° = 90° + k. 360° b. Interval fungsi naike I f'(x) >0) f'@) = = 2 as (2x+60) >0 2852X < 360° 5 2X = 30° tk. 360° x = 1 tk. 180° 2 > K-0 -0 X = 15' :>k: 140 x = 1959 + - + + 0° وا 105 1959 285 360° Pahamify
3) y= sin x + OS X, Ò EX €360° Ditanya: interval fungu turun y': = Los X-SM X # Syarat fungsi turun y y'<i » WsX - SIN X <o 0 = los X - sinx + t Sinx = Wsx Sinx D 45 225 360° cosx 5 & interval y turun 45° <x<225° 2 fan x=tan 450 X = 450 tk. 180° ok=0 - X=45° > k=1 X = 225' = Pahamify 
A f(x) X₂ UiS -X sinex a 0 = Xc X = 1 0 X° D. 50 Oʻ = x < 360° # Subtitusi nilai - Millal x ke f(x) = an’x < Ditanya koordinat titik stasioner flo)=0 > X = 90' of (90°) = Syarat! f(x)=0 -> x = 180° -> $(180) 0 f'(x) - 2 sin x.asx f(360) ={{x,y)(0,0),(507,1) , (10%,0), (270".), # X=0tk.360° # 2X= 180°tk. 360° (360º,0)) 1 = (otzita ,otz exc. a 2 :> X > 360° - = 0 = Sin 2x 0 sin 2x = sin o 5 .081.77.06- X ,lb=XG 0=716 x - k. 180° sk=0 +x=0 ?K= 1 + x = 180° 7 k= 2 X=360° o sk=10x=270° Pahamify 
+ lo 0<x<15 atau 105°<x< 1959 atau . a ) 3 f(x) = sin (2X+60), 0 < x < 360° a. absis fitik stasioner (f '(x)=0) + 2x + 60°= -90°+k. 360° a. °= 2x = -150° +1.360° f'(x) - 2 cos (2X+60) X = -75°4k, 180° 2 cos (2x+60) = 0 -> k = 1 x = 105 as (2x + 60°) = 0 »k=20 X = 285° cos (2x+60) = cos go 3. X= { IS', 105°, 1959, 285"} # 2x+60° = 90° + k. 360° b. Interval fungsi naike I f'(x) >0) f'@) = = 2 as (2x+60) >0 2852X < 360° 5 2X = 30° tk. 360° x = 1 tk. 180° 2 > K-0 -0 X = 15' :>k: 140 x = 1959 + - + + 0° وا 105 1959 285 360° Pahamify
3) y= sin x + OS X, Ò EX €360° Ditanya: interval fungu turun y': = Los X-SM X # Syarat fungsi turun y y'<i » WsX - SIN X <o 0 = los X - sinx + t Sinx = Wsx Sinx D 45 225 360° cosx 5 & interval y turun 45° <x<225° 2 fan x=tan 450 X = 450 tk. 180° ok=0 - X=45° > k=1 X = 225' = Pahamify 
A f(x) X₂ UiS -X sinex a 0 = Xc X = 1 0 X° D. 50 Oʻ = x < 360° # Subtitusi nilai - Millal x ke f(x) = an’x < Ditanya koordinat titik stasioner flo)=0 > X = 90' of (90°) = Syarat! f(x)=0 -> x = 180° -> $(180) 0 f'(x) - 2 sin x.asx f(360) ={{x,y)(0,0),(507,1) , (10%,0), (270".), # X=0tk.360° # 2X= 180°tk. 360° (360º,0)) 1 = (otzita ,otz exc. a 2 :> X > 360° - = 0 = Sin 2x 0 sin 2x = sin o 5 .081.77.06- X ,lb=XG 0=716 x - k. 180° sk=0 +x=0 ?K= 1 + x = 180° 7 k= 2 X=360° o sk=10x=270° Pahamify 
+ lo 0<x<15 atau 105°<x< 1959 atau . a ) 3 f(x) = sin (2X+60), 0 < x < 360° a. absis fitik stasioner (f '(x)=0) + 2x + 60°= -90°+k. 360° a. °= 2x = -150° +1.360° f'(x) - 2 cos (2X+60) X = -75°4k, 180° 2 cos (2x+60) = 0 -> k = 1 x = 105 as (2x + 60°) = 0 »k=20 X = 285° cos (2x+60) = cos go 3. X= { IS', 105°, 1959, 285"} # 2x+60° = 90° + k. 360° b. Interval fungsi naike I f'(x) >0) f'@) = = 2 as (2x+60) >0 2852X < 360° 5 2X = 30° tk. 360° x = 1 tk. 180° 2 > K-0 -0 X = 15' :>k: 140 x = 1959 + - + + 0° وا 105 1959 285 360° Pahamify
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