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Diketahui f ( x ) = x 2 + 4 x + 4 dan g ( x ) = x 2 − 4 x + 4 . Rumus f ( g ( x ) ) ​ + g ( f ( x ) ) ​ = ....

Diketahui  dan . Rumus  

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Diketahui dan . Rumus kita sederhanakan dulu fungsinya, = = kemudian tentukan dulu fungsi komposisinya: jadi, jawabannya C

Diketahui  dan . Rumus  

kita sederhanakan dulu fungsinya,

 = begin mathsize 14px style open parentheses x plus 2 close parentheses squared end style 

 = begin mathsize 14px style open parentheses x minus 2 close parentheses squared end style 

kemudian tentukan dulu fungsi komposisinya:

size 14px f size 14px left parenthesis size 14px g size 14px left parenthesis size 14px x size 14px right parenthesis size 14px right parenthesis size 14px equals size 14px f size 14px left parenthesis size 14px left parenthesis size 14px x size 14px minus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px right parenthesis size 14px equals size 14px left parenthesis size 14px left parenthesis size 14px x size 14px minus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px equals begin mathsize 14px style left parenthesis x squared minus 4 x plus 4 plus 2 right parenthesis end style to the power of size 14px 2 size 14px equals begin mathsize 14px style left parenthesis x squared minus 4 x plus 6 right parenthesis end style to the power of size 14px 2 size 14px g size 14px left parenthesis size 14px f size 14px left parenthesis size 14px x size 14px right parenthesis size 14px right parenthesis size 14px equals size 14px g size 14px left parenthesis size 14px left parenthesis size 14px x size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px right parenthesis size 14px equals size 14px left parenthesis size 14px left parenthesis size 14px x size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px minus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px equals begin mathsize 14px style left parenthesis x squared plus 4 x plus 4 minus 2 right parenthesis end style to the power of size 14px 2 size 14px equals size 14px left parenthesis size 14px x to the power of size 14px 2 size 14px plus size 14px 4 size 14px x size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 size 14px space  square root of size 14px f size 14px left parenthesis size 14px g size 14px left parenthesis size 14px x size 14px right parenthesis size 14px right parenthesis end root size 14px plus square root of size 14px g size 14px left parenthesis size 14px f size 14px left parenthesis size 14px x size 14px right parenthesis size 14px right parenthesis end root size 14px equals square root of begin mathsize 14px style left parenthesis x squared minus 4 x plus 6 right parenthesis squared end style end root size 14px plus square root of size 14px left parenthesis size 14px x to the power of size 14px 2 size 14px plus size 14px 4 size 14px x size 14px plus size 14px 2 size 14px right parenthesis to the power of size 14px 2 end root size 14px equals size 14px x to the power of size 14px 2 size 14px minus size 14px 4 size 14px x size 14px plus size 14px 6 size 14px plus size 14px x to the power of size 14px 2 size 14px plus size 14px 4 size 14px x size 14px plus size 14px 2 size 14px equals size 14px 2 size 14px x to the power of size 14px 2 size 14px plus size 14px 8

jadi, jawabannya C

 

 

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Jika f(x) = x − 2 dan g ( x ) = x + 2 , fungsi { ( f ∘ g ) ( x ) } ⋅ { ( g ∘ f ( x ) ) } = ....

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