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

Answer:

[tex]f^{-1}\,\circ \,g(5) = 10[/tex]

Step-by-step explanation:

Let [tex]f(x+7) = 3\cdot x -6[/tex] and [tex]g(2\cdot x +1) = x^{2}-1[/tex], we proceed to derive [tex]f(x)[/tex] and [tex]g(x)[/tex] by algebraic means:

(i)  [tex]f(x+7) = 3\cdot x -6[/tex]

1) [tex]f(x+7) = 3\cdot x -6[/tex] Given

2) [tex]f(x+7) = 3\cdot (x+0) - 6[/tex] Modulative property

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

4) [tex]f(x+7) = 3\cdot (x+7) +3\cdot (-7)-6[/tex] Distributive property

5) [tex]f(x+7) = 3\cdot (x+7) -21-6[/tex]   [tex]a\cdot (-b) = -a\cdot b[/tex]

6) [tex]f(x+7) = 3\cdot (x+7) -27[/tex] Definition of subtraction

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

(ii) [tex]g(2\cdot x + 1) = x^{2}-1[/tex]

1) [tex]g(2\cdot x + 1) = x^{2}-1[/tex] Given

2) [tex]g(2\cdot x + 1) = (x\cdot 1)^{2}-1[/tex] Modulative property

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

4) [tex]g(2\cdot x +1) = (2\cdot x)^{2}\cdot 2^{-2}-1[/tex]   [tex]a^{c}\cdot b^{c}[/tex]/[tex](a^{b})^{c} = a^{b\cdot c}[/tex]

5) [tex]g(2\cdot x + 1) = \frac{(2\cdot x)^{2}}{4}-1[/tex] Definitions of division and power

6) [tex]g(2\cdot x + 1) = \frac{(2\cdot x + 0)^{2}}{4} -1[/tex] Modulative property

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

8) [tex]g(2\cdot x + 1) = \frac{(2\cdot x + 1)^{2}+2\cdot (2\cdot x + 1)\cdot (-1)+(-1)^{2}}{4} -1[/tex] Perfect square trinomial

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

10) [tex]g(x) = \frac{x^{2}}{4}-\frac{2\cdot x}{4} + \frac{1}{4}-1[/tex] Composition of functions/[tex]a\cdot (-b) = -a\cdot b[/tex]

11) [tex]g(x) = \frac{x^{2}}{4}-\frac{x}{2}-\frac{3}{4}[/tex] Definitions of division and subtraction/Result

Now we find the inverse of [tex]f(x)[/tex]:

1) [tex]f = 3\cdot x - 27[/tex] Given

2) [tex]f + 27 = (3\cdot x - 27)+27[/tex] Compatibility with addition

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

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

5) [tex](f+27) \cdot 3^{-1} = (3\cdot 3^{-1})\cdot x[/tex] Compatibility with multiplication/Commutative and associative properties

6) [tex](f+27)\cdot 3^{-1} = x[/tex] Existence of multiplicative inverse/Modulative property

7) [tex]f^{-1} (x) = \frac{x+27}{3}[/tex] Symmetrical property/Notation/Result

Finally, we proceed to calculate [tex]f^{-1}\,\circ \, g (5)[/tex]:

1) [tex]f^{-1} (x) = \frac{x+27}{3}[/tex], [tex]g(x) = \frac{x^{2}}{4}-\frac{x}{2}-\frac{3}{4}[/tex]  Given

2) [tex]f^{-1}\,\circ\, g(x) = \frac{\frac{x^{2}}{4}-\frac{x}{2}-\frac{3}{4}+27}{3}[/tex] Composition of functions

3) [tex]f^{-1}\,\circ \,g(5) = 10[/tex] Result

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