Respuesta :

VictorZ
Hi there!

• g(x) = 6x - 3

Then,
if g(x) = y :-

y = 6x - 3

⇒6x = y + 3

⇒x = [tex]\dfrac{\text{y} + 3}{6}[/tex]

According to th' question :-

g⁻¹(x) = [tex]\dfrac{\text{x} + 3}{6}[/tex]

g⁻¹(9) = [tex]\dfrac{9 + 3}{6}[/tex]

g⁻¹(9) = [tex]\dfrac{12}{6}[/tex]

g⁻¹(9) = 2

Hence,
The required answer is g⁻¹(9) = 2

~ Hope it helps!

g(x) = 6x - 3


Let g(x) = y:

y = 6x - 3


Make x the subject:

y = 6x - 3

6x = y + 3

[tex] x = \dfrac{y + 3}{6} [/tex]


Replace y with x:

[tex] g^{-1}(x) = \dfrac{x + 3}{6} [/tex]


Find g⁻¹(9):

[tex] g^{-1}(x) = \cfrac{x + 3}{6} [/tex]

[tex] g^{-1}(9) = \cfrac{9 + 3}{6} [/tex] 

[tex] g^{-1}(9) = 2 [/tex] 

Answer: 2



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