10. ( f o g ) (x) = f [ g(x) ]
x² - 4 = f( x + 3 )
Misal : x + 3 = R
x = R - 3
x² - 4 = f( x + 3 )
( R - 3 )² - 4 = f(R)
R² - 6R + 9 - 4 = f(R)
R² - 6R + 5 = f(R)
f(x) = x² - 6x + 5
f(x-2) = (x-2)² - 6(x-2) + 5
= x² - 4x + 4 - 6x + 12 + 5
= x² - 10x + 21 ( C. )
11. ( f o g ) (x) = f [ g(x) ]
4x² + 20x + 23 = f ( 2x + 5 )
Misal : 2x + 5 = R
2x = R - 5
x = ½( R - 5 )
4x² + 20x + 23 = f( 2x + 5 )
4[ ½( R - 5 ) ]² + 20[ ½( R - 5 ) ] + 23 = f(R)
4[ ¼( R² - 10R + 25 ) ] + 10( R - 5 ) + 23 = f(R)
R² - 10R + 25 + 10R - 50 + 23 = f(R)
R² - 2 = f(R)
f(x) = x² - 2 ( A. )
12. ( g o f ) (x) = g[ f(x) ]
2x² + 4x + 5 = 2f(x) + 3
2x² + 4x + 5 - 3 = 2f(x)
2x² + 4x + 2 = 2f(x)
f(x) = x² + 2x + 1 ( A. )
13. ( f o g ) (x) = f [ g(x) ]
[tex] = 3( \frac{4x - 5}{2x + 1} ) + 4 \\ = \frac{12x - 15}{2x + 1} + 4 \\ = \frac{12x - 15 +4(2x + 1) }{2x + 1} \\ = \frac{12x - 15 + 8x + 4}{2x + 1} \\ = \frac{20x - 11}{2x + 1} [/tex]
[tex](fog)(x) = y \\ \frac{20x - 11}{2x + 1} = y [/tex]
20x - 11 = y ( 2x + 1 )
20x - 11 = 2yx + y
20x - 2yx = y + 11
( 20 - 2y ) x = y + 11
[tex]x = \frac{ y + 11}{20 + 2y} \\ (fog) {}^{ - 1} (x) = \frac{ x + 11}{20 + 2x} [/tex]
[answer.2.content]
