Question

What is the value of g^-1(9)

Answer 1

g(x) = 6x - 3

Let g(x) = y:

y = 6x - 3

Make x the subject:

y = 6x - 3

6x = y + 3

$x = \dfrac{y + 3}{6}$

Replace y with x:

$g^{-1}(x) = \dfrac{x + 3}{6}$

Find g⁻¹(9):

$g^{-1}(x) = \cfrac{x + 3}{6}$

$g^{-1}(9) = \cfrac{9 + 3}{6}$ 

$g^{-1}(9) = 2$ 

Answer: 2

Answer 2

Hi there!

• g(x) = 6x - 3

Then,
if g(x) = y :-

y = 6x - 3

⇒6x = y + 3

⇒x = $\dfrac{\text{y} + 3}{6}$

According to th' question :-

g⁻¹(x) = $\dfrac{\text{x} + 3}{6}$

g⁻¹(9) = $\dfrac{9 + 3}{6}$

g⁻¹(9) = $\dfrac{12}{6}$

g⁻¹(9) = 2

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

~ Hope it helps!