Question-8387

Question Number 8387 by arinto27 last updated on 09/Oct/16

Answered by ridwan balatif last updated on 10/Oct/16

1 . f (x) = 5 x + 1

y = 5 x + 1

x = \frac{y - 1}{5}

f^{- 1} (x) = \frac{x - 1}{5} \Leftrightarrow f^{- 1} (a) = \frac{a - 1}{5} \Leftrightarrow 15 = a - 1 \Leftrightarrow a = 16

2 . f (x) = x + 3 dan g (x) = x^{2} - 2 x - 1

(f o g) (x) = f (g (x))

= f (x^{2} - 2 x - 1)

= (x^{2} - 2 x - 1) + 3

= x^{2} - 2 x + 2

3 . f (x) = ax + 3

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

f^{- 1} (f^{- 1} (x)) = ax + 3

f^{- 1} (f^{- 1} (9)) = a . 9 + 3

3 = 9 a + 3

a = 0

a^{2} + a + 1 = 0 + 0 + 1 = 1

Commented by 123456 last updated on 10/Oct/16

1 : f (x) = 5 x + 1, f^{- 1} (a) = 3

f^{- 1} (a) = 3

f (f^{- 1} (a)) = f (3) = 3 \times 5 + 1 = 16

a = 16

Commented by 123456 last updated on 10/Oct/16

3 : f (x) = a x + 3, f^{- 1} (f^{- 1} (9)) = 3

f^{- 1} (f^{- 1} (9)) = 3

f (f^{- 1} (f^{- 1} (9))) = f (3) = 3 a + 3 = 3 (a + 1)

f^{- 1} (9) = 3 (a + 1)

f (f^{- 1} (9)) = f (3 (a + 1)) = 3 (a + 1) a + 3

9 = 3 ((a + 1) a + 1) = 3 (a^{2} + a + 1)

a^{2} + a + 1 = 3

Commented by sandy_suhendra last updated on 10/Oct/16

3) f^{- 1} [f^{- 1} (x)] = \frac{\frac{x - 3}{a} - 3}{a} = \frac{x - 3 - 3 a}{a^{2}}

f^{- 1} [f^{- 1} (9)] = \frac{9 - 3 - 3 a}{a^{2}} = 3

{3 a}^{2} = 6 - 3 a

{3 a}^{2} + 3 a = 6

a^{2} + a = 2

a^{2} + a + 1 = 2 + 1 = 3