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liyy
Student XII IPA
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Putra Gema Nusa
Rockstar Teacher GURU SMA
b. b. Interval naik = f'(x) >0 f(x) = 3 as (x + " ,0 5x+ 20 f: sin 4 -3 sin (x+4) >0 135 315 > 5 sin Ixt misalkan : sin (x+ I)=0 » > x=0 sin (0+45°) <0 Sin 49° <o X=135 X=315° Đ 12 12 co X : Interval nack B8 < X <315 Dahamify 
= 3.1=3 > f(x)= 3 aos (x + 1) .05X621 a. titik stasioner f'(x) = 0 # X=315° f'(x) = - 3 Sın (x+ I)I | f(315) =3 as (3159 +45) = 3 cos 360° 0 =-3 sin(x + F sin (x+4)= sino # x = 135° » X, + 45° =0tk. 360° ° f (1959) = 3 cos (1850 + 45") X, = -45° tk. 360 -3. as 180° -> k = 1 + X. = 315° = 3.-|=-3 » X₂ + 45° = (180°-0) tk. 360° fitlk stasionernya adalah X₂ = 1350 tk. 360° -3 sk=0 <D X₂ = 135 Dahamify s _18
b. b. Interval naik = f'(x) >0 f(x) = 3 as (x + " ,0 5x+ 20 f: sin 4 -3 sin (x+4) >0 135 315 > 5 sin Ixt misalkan : sin (x+ I)=0 » > x=0 sin (0+45°) <0 Sin 49° <o X=135 X=315° Đ 12 12 co X : Interval nack B8 < X <315 Dahamify 
= 3.1=3 > f(x)= 3 aos (x + 1) .05X621 a. titik stasioner f'(x) = 0 # X=315° f'(x) = - 3 Sın (x+ I)I | f(315) =3 as (3159 +45) = 3 cos 360° 0 =-3 sin(x + F sin (x+4)= sino # x = 135° » X, + 45° =0tk. 360° ° f (1959) = 3 cos (1850 + 45") X, = -45° tk. 360 -3. as 180° -> k = 1 + X. = 315° = 3.-|=-3 » X₂ + 45° = (180°-0) tk. 360° fitlk stasionernya adalah X₂ = 1350 tk. 360° -3 sk=0 <D X₂ = 135 Dahamify s _18
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