Tolong di bantu jawab ya KK!...

Tolong di bantu jawab ya KK!…

1. ( g o f )(x) = g[ f(x) ]

= g( 2x + 1 )

= 5( 2x + 1 )²

= 5( 4x² + 4x + 1 )

= 20x² + 20x + 5

2. ( f o g )(x) = f[ g(x) ]

6x + 4 = 2g(x)

g(x) = ( 6x + 4 )/2

g(x) = 3x + 2

3. ( f o g )(x) = f[ g(x) ]

= f(4x)

= 5(4x)² + 1

= 5.16x² + 1

= 80x² + 1

( f o g )(2) = 80.2² + 1

= 80.4 + 1

= 320 + 1 = 321

4. ( f o g o h )(x) = f o g[ h(x) ]

= f o g(3x)

= f[ 5(3x) ]

= f(15x)

= 2(15x) + 1

= 30x + 1

( f o g o h )(1) = 30.1 + 1 = 31

5. f(x) = y

3 – 3x = y

-3x = y – 3

$x = frac{y - 3}{ - 3} \ f {}^{ - 1} (x) = frac{x - 3}{ - 3}$

6. f(x) = y

$frac{3x - 2}{5x + 8} = y$

3x + 2 = y( 5x + 8 )

3x + 2 = 5yx + 8y

3x – 5yx = 8y – 2

( 3 – 5y ) x = 8y – 2

$x = frac{8y - 2}{3 - 5y} \ f {}^{ - 1}(x) = frac{8x - 2}{3 - 5x}$

7. ( g o f )(x) = g[ f(x) ]

= g( x – 3 )

= 2( x – 3 ) + 4

= 2x – 6 + 4

= 2x – 2

( g o f )(x) = y

2x – 2 = y

2x = y + 2

$x = frac{y + 2}{2} \ (gof) {}^{ - 1} (x) = frac{y + 2}{2} \ (gof) {}^{ - 1} (4) = frac{4 + 2}{2} \ = frac{6}{2} = 3$

8. ( f o g )(x) = f[ g(x) ]

$frac{3x - 4}{2x} = frac{5}{g(x)}$

g(x)( 3x – 4 ) = 2x.5

g(x)( 3x – 4 ) = 10x

$g(x) = frac{10x}{3x - 4}$

g(x) = y

$frac{10x}{3x - 4} = y$

10x = y( 3x – 4 )

10x = 3yx – 4y

10x – 3yx = -4y

( 10 – 3y ) x = -4y

$x = frac{ - 4y}{10 - 3y} \ g {}^{ - 1}(x) = frac{ - 4x}{10 - 3x} \ g {}^{ - 1} (2) = frac{ - 4(2)}{10 - 3(2)} \ = frac{ - 8}{10 - 6} \ = frac{ - 8}{4} = - 2$

9. f(x) = y

$( sqrt{x + 5 }) {}^{2} =( y) {}^{2}$

x + 5 = y²

x = y² – 5

f⁻¹(x) = x² – 5

10. f(x) = y

x² – 3x + 9 = y

a = 1, b = -3, c = 9

$f {}^{ - 1} (x) = frac{ - b + - sqrt{b {}^{2} - 4a(c - y) } }{2a} \ = frac{3 + - sqrt{( - 3) {}^{2} - 4.1(9 - y) } }{2.1} \ = frac{3 + - sqrt{9 -36 + 4y} }{2} \ f {}^{ - 1} (x) = frac{3 + - sqrt{4y - 27} }{2}$