FUngsi KOMposisi

Penjelasan dengan langkah-langkah:

1/ f(x)= x³ + 4  dan g(x) = 2 sin x
nilai  fog(¹/₂π)=
= f{g(¹/₂π)}
= f{2 sin /₂π}
= f {2(1)}
= f(2)
= 2³+ 4
= 12

2. g(x)=  3x + 7
gof(x) = 15x²- 6x +19
g{f(x)}  = 15x² - 6x+19
3. f(x) + 7 = 15x² - 6x +19
3. f(x) = 15x² - 6x +19- 7
3. f(x) = 15x² - 6x +12
f(x) =  5x² - 2x + 4

3. f(x) = 2x + 3
g(x)= 3x- 1

{fog}(x-4) = f(x) + 2.g(x)
f{3(x-4)-1 } = 2x + 3 + 2 (3x  -1)
f(3x-12 -1)  = 2x + 3 + 6x - 2
f(3x-13) =  8x + 1
2(3x-13) + 3 =  8x + 1
6x - 26  + 3 = 8x + 1
6x - 23 = 8x + 1
6x - 8x = 1 +23
2x = 24
x = 12

4. f(x) = x + 1
{fog}(x) = 3x² + 4
f{g(x)} =  3x²+ 1
g(x) + 1 =  3x² + 1
g(x)=  3x²
g(4) = 3.4² = 3 (16)
g(4) = 48