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Inga [223]
3 years ago
15

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

Mathematics
1 answer:
Murljashka [212]3 years ago
3 0

Answer:

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

Step-by-step 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) = \frac{(2\cdot x)^{2}}{4}-1 Definitions of division and power

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

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

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

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

11) g(x) = \frac{x^{2}}{4}-\frac{x}{2}-\frac{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) = \frac{x+27}{3} Symmetrical property/Notation/Result

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

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

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

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

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