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fifah
Student XII IPA
TI Diketahui g(x) = -2 cos (2x + 3) untuk 0<x<211. Fungsi g cekung ke atas pada interval ....
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Putra Gema Nusa
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Pahamify + U .098 X 3.0tr nav .08) 3 X5 206 .19€ Itz 2081 B6 0 + + - cos (2x +E) >0 
I 2 7 + f"(x) > 0 2x = k. 360° X x - k. 180° =k. aos E d g(x) = – 2 as (2x + ) ,0 < x < 211 2 X o Ditanya: interval fungsi cekung ke atas as (2x + ) >cos 90° → = 90° Ingat! # 2X + 90o = 90° tk. 360° Cekung keatas, maka g(x) = -2 213 (2x'+1) ( g'(x) = -2.(- 2 sin (2x + +7) 4'(x) = 4 sm (2x + y) g'(x) = 4.2 as [2x +) # 2x +90° =-90° tk. 360° ga(w) = 9 as(2x + }) + ) 8 os (2x++) > 0 Pahamify as (2x+E) >0 > is k=0 &&0 k=1 DX 7180° ak= 2 x € 360° > 5 2x = -180° + K. 360° x =-goo tk. 180° sk=I LDX E90) -> k= 20 X 7270°
Pahamify + U .098 X 3.0tr nav .08) 3 X5 206 .19€ Itz 2081 B6 0 + + - cos (2x +E) >0 
I 2 7 + f"(x) > 0 2x = k. 360° X x - k. 180° =k. aos E d g(x) = – 2 as (2x + ) ,0 < x < 211 2 X o Ditanya: interval fungsi cekung ke atas as (2x + ) >cos 90° → = 90° Ingat! # 2X + 90o = 90° tk. 360° Cekung keatas, maka g(x) = -2 213 (2x'+1) ( g'(x) = -2.(- 2 sin (2x + +7) 4'(x) = 4 sm (2x + y) g'(x) = 4.2 as [2x +) # 2x +90° =-90° tk. 360° ga(w) = 9 as(2x + }) + ) 8 os (2x++) > 0 Pahamify as (2x+E) >0 > is k=0 &&0 k=1 DX 7180° ak= 2 x € 360° > 5 2x = -180° + K. 360° x =-goo tk. 180° sk=I LDX E90) -> k= 20 X 7270°
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