Question

- f:R → R
f(x + 7) = 3x - 6
g(2x + 1) = x2-1
= (f-1og)(5) = ?

Answer 1

`f^(-1)\,\circ \,g(5) = 10`

Explanation:

Let
`f(x+7) = 3\cdot x -6` and
`g(2\cdot x +1) = x^(2)-1`, we proceed to derive
`f(x)` and
`g(x)` by algebraic means:

(i)
`f(x+7) = 3\cdot x -6`

1)
`f(x+7) = 3\cdot x -6` Given

2)
`f(x+7) = 3\cdot (x+0) - 6` Modulative property

3)
`f(x+7) = 3\cdot [(x+7) +(-7)]-6` Existence of additive inverse/Associative property

4)
`f(x+7) = 3\cdot (x+7) +3\cdot (-7)-6` Distributive property

5)
`f(x+7) = 3\cdot (x+7) -21-6`
`a\cdot (-b) = -a\cdot b`

6)
`f(x+7) = 3\cdot (x+7) -27` Definition of subtraction

7)
`f(x) = 3\cdot x - 27` Composition of functions/Result

(ii)
`g(2\cdot x + 1) = x^(2)-1`

1)
`g(2\cdot x + 1) = x^(2)-1` Given

2)
`g(2\cdot x + 1) = (x\cdot 1)^(2)-1` Modulative property

3)
`g(2\cdot x +1) = [(2\cdot x)\cdot 2^(-1)]^(2)-1` Existence of additive inverse/Commutative and associative properties

4)
`g(2\cdot x +1) = (2\cdot x)^(2)\cdot 2^(-2)-1`
`a^(c)\cdot b^(c)`/
`(a^(b))^(c) = a^(b\cdot c)`

5)
`g(2\cdot x + 1) = ((2\cdot x)^(2))/(4)-1` Definitions of division and power

6)
`g(2\cdot x + 1) = ((2\cdot x + 0)^(2))/(4) -1` Modulative property

7)
`g(2\cdot x +1) = ([(2\cdot x + 1)+(-1)]^(2))/(4) -1` Existence of additive inverse/Associative property

8)
`g(2\cdot x + 1) = ((2\cdot x + 1)^(2)+2\cdot (2\cdot x + 1)\cdot (-1)+(-1)^(2))/(4) -1` Perfect square trinomial

9)
`g(2\cdot x + 1) = ((2\cdot x + 1)^(2))/(4)+([2\cdot (-1)]\cdot (2\cdot x + 1))/(4) +((-1)^(2))/(4)-1` Addition of homogeneous fractions.

10)
`g(x) = (x^(2))/(4)-(2\cdot x)/(4) + (1)/(4)-1` Composition of functions/
`a\cdot (-b) = -a\cdot b`

11)
`g(x) = (x^(2))/(4)-(x)/(2)-(3)/(4)` Definitions of division and subtraction/Result

Now we find the inverse of
`f(x)`:

1)
`f = 3\cdot x - 27` Given

2)
`f + 27 = (3\cdot x - 27)+27` Compatibility with addition

3)
`f+ 27 = 3\cdot x +[27+(-27)]` Definition of substraction/Commutative and associative properties

4)
`f+27 = 3\cdot x` Existence of additive inverse/Modulative property

5)
`(f+27) \cdot 3^(-1) = (3\cdot 3^(-1))\cdot x` Compatibility with multiplication/Commutative and associative properties

6)
`(f+27)\cdot 3^(-1) = x` Existence of multiplicative inverse/Modulative property

7)
`f^(-1) (x) = (x+27)/(3)` Symmetrical property/Notation/Result

Finally, we proceed to calculate
`f^(-1)\,\circ \, g (5)`:

1)
`f^(-1) (x) = (x+27)/(3)`,
`g(x) = (x^(2))/(4)-(x)/(2)-(3)/(4)` Given

2)
`f^(-1)\,\circ\, g(x) = ((x^(2))/(4)-(x)/(2)-(3)/(4)+27)/(3)` Composition of functions

3)
`f^(-1)\,\circ \,g(5) = 10` Result