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Student XII IPA
) 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
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3) ľ < X < 360° 1 y= sin x + OS X Ditanya: interval y': = Los X-shX turun issunt # Syarat fungsi turun 0 > X VIS – XSCO a ezin o los X - sinx 3 Sinx = Wsx t t Sinx 0. 45° 225 ,99€ | = 5 cosx U fan x = tan 45 x = 450 + k. 180* tk° ok=0 X= 45° > Kol D X = 225' o interval y turun 45° < x < 225° ° x x o Pahamify 
C 2) f(x) = sin (2x +60), 0 < x < 360° x < a. absis' fithik stasioner (f'(x)=0) + 2x + 60°= -90°+K.360° f'(x) = 2 cos (2X + 60') 2 cos (2x+60) = 0 as (2x + 60°) = 0 as (2x +60) = cos go :. X = {is”, 105°, 195°, 285"} # 2x+60° = 90° + k. 360° b. Interval fungsi Mare | f'(x) >0) 2 > le o ex<15 atau 105° <x< 1950 atau 285 2X < 360° a 2x = -150° tk. 360° x = -75 tk. 180° -> k = 1 0 X = 1050 "k=20X=285° S 45 2X = 30° tk. 360° x = 14 tk. 180° f'@) = 2 as (2x+60' ) >0 © -> k=0 -DX=15' > kilox=195° + + + Dahamify Dº اکی 105 1950 285 360° 
A Яп y f(x) = sin²x » X=0 -D flo)=0 -> = -> °. O'< x < 360 # Subtitusi nilai-nilai x ke f®) = sin’x os X=# X Ditanya koordinat titik stasioner =... x = 90° of (90°) = 1 Syarat! f'(x)=0 > X = 180° 1180) >0 of( f'W) - 2 sin x.asx x -> X = 270' +_f(270) = 1 2{(x,y)(0,0),1909 , 1) (180",") (270" 1) (3601,0)) 2 :> X = 360° 0 f(360°) =0 # 는 n 0 = sin 2x sin 2x sin o # 2X=0 tk. 360° # 2X= 180°tk.360° x = k. 180° x = 90'tk. 180° sk=0 X=0 >k=0 ox=90° >kal » X= 180° sk=108=270° 7 k= 2 X=360° Đ Pahamify
3) ľ < X < 360° 1 y= sin x + OS X Ditanya: interval y': = Los X-shX turun issunt # Syarat fungsi turun 0 > X VIS – XSCO a ezin o los X - sinx 3 Sinx = Wsx t t Sinx 0. 45° 225 ,99€ | = 5 cosx U fan x = tan 45 x = 450 + k. 180* tk° ok=0 X= 45° > Kol D X = 225' o interval y turun 45° < x < 225° ° x x o Pahamify 
C 2) f(x) = sin (2x +60), 0 < x < 360° x < a. absis' fithik stasioner (f'(x)=0) + 2x + 60°= -90°+K.360° f'(x) = 2 cos (2X + 60') 2 cos (2x+60) = 0 as (2x + 60°) = 0 as (2x +60) = cos go :. X = {is”, 105°, 195°, 285"} # 2x+60° = 90° + k. 360° b. Interval fungsi Mare | f'(x) >0) 2 > le o ex<15 atau 105° <x< 1950 atau 285 2X < 360° a 2x = -150° tk. 360° x = -75 tk. 180° -> k = 1 0 X = 1050 "k=20X=285° S 45 2X = 30° tk. 360° x = 14 tk. 180° f'@) = 2 as (2x+60' ) >0 © -> k=0 -DX=15' > kilox=195° + + + Dahamify Dº اکی 105 1950 285 360° 
A Яп y f(x) = sin²x » X=0 -D flo)=0 -> = -> °. O'< x < 360 # Subtitusi nilai-nilai x ke f®) = sin’x os X=# X Ditanya koordinat titik stasioner =... x = 90° of (90°) = 1 Syarat! f'(x)=0 > X = 180° 1180) >0 of( f'W) - 2 sin x.asx x -> X = 270' +_f(270) = 1 2{(x,y)(0,0),1909 , 1) (180",") (270" 1) (3601,0)) 2 :> X = 360° 0 f(360°) =0 # 는 n 0 = sin 2x sin 2x sin o # 2X=0 tk. 360° # 2X= 180°tk.360° x = k. 180° x = 90'tk. 180° sk=0 X=0 >k=0 ox=90° >kal » X= 180° sk=108=270° 7 k= 2 X=360° Đ Pahamify
3) ľ < X < 360° 1 y= sin x + OS X Ditanya: interval y': = Los X-shX turun issunt # Syarat fungsi turun 0 > X VIS – XSCO a ezin o los X - sinx 3 Sinx = Wsx t t Sinx 0. 45° 225 ,99€ | = 5 cosx U fan x = tan 45 x = 450 + k. 180* tk° ok=0 X= 45° > Kol D X = 225' o interval y turun 45° < x < 225° ° x x o Pahamify 
C 2) f(x) = sin (2x +60), 0 < x < 360° x < a. absis' fithik stasioner (f'(x)=0) + 2x + 60°= -90°+K.360° f'(x) = 2 cos (2X + 60') 2 cos (2x+60) = 0 as (2x + 60°) = 0 as (2x +60) = cos go :. X = {is”, 105°, 195°, 285"} # 2x+60° = 90° + k. 360° b. Interval fungsi Mare | f'(x) >0) 2 > le o ex<15 atau 105° <x< 1950 atau 285 2X < 360° a 2x = -150° tk. 360° x = -75 tk. 180° -> k = 1 0 X = 1050 "k=20X=285° S 45 2X = 30° tk. 360° x = 14 tk. 180° f'@) = 2 as (2x+60' ) >0 © -> k=0 -DX=15' > kilox=195° + + + Dahamify Dº اکی 105 1950 285 360° 
A Яп y f(x) = sin²x » X=0 -D flo)=0 -> = -> °. O'< x < 360 # Subtitusi nilai-nilai x ke f®) = sin’x os X=# X Ditanya koordinat titik stasioner =... x = 90° of (90°) = 1 Syarat! f'(x)=0 > X = 180° 1180) >0 of( f'W) - 2 sin x.asx x -> X = 270' +_f(270) = 1 2{(x,y)(0,0),1909 , 1) (180",") (270" 1) (3601,0)) 2 :> X = 360° 0 f(360°) =0 # 는 n 0 = sin 2x sin 2x sin o # 2X=0 tk. 360° # 2X= 180°tk.360° x = k. 180° x = 90'tk. 180° sk=0 X=0 >k=0 ox=90° >kal » X= 180° sk=108=270° 7 k= 2 X=360° Đ Pahamify
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