To determine if the function [tex]\( g(x) = \frac{x+1}{6} \)[/tex] has an inverse, we need to check if it is one-to-one. A function has an inverse if and only if it is bijective, which means it must be both injective (one-to-one) and surjective (onto).
For the function to be one-to-one, its derivative must be non-zero for all [tex]\( x \in \mathbb{R} \)[/tex]. Let's find the derivative of [tex]\( g(x) \)[/tex]:
[tex]\[ g'(x) = \frac{d}{dx}\left(\frac{x+1}{6}\right) \][/tex]
Differentiating [tex]\( \frac{x+1}{6} \)[/tex], we get:
[tex]\[ g'(x) = \frac{1}{6} \][/tex]
Since the derivative [tex]\( g'(x) = \frac{1}{6} \)[/tex] is a constant and is non-zero for all [tex]\( x \)[/tex], the function [tex]\( g(x) \)[/tex] is one-to-one. Therefore, [tex]\( g(x) \)[/tex] has an inverse.
To find the inverse function [tex]\( g^{-1}(x) \)[/tex], we will solve the equation [tex]\( g(y) = x \)[/tex] for [tex]\( y \)[/tex]:
[tex]\[ x = \frac{y+1}{6} \][/tex]
To isolate [tex]\( y \)[/tex], we multiply both sides by 6:
[tex]\[ 6x = y + 1 \][/tex]
Next, we solve for [tex]\( y \)[/tex] by subtracting 1 from both sides:
[tex]\[ y = 6x - 1 \][/tex]
Therefore, the inverse function is:
[tex]\[ g^{-1}(x) = 6x - 1 \][/tex]
Summarizing, yes, [tex]\( g \)[/tex] does have an inverse, and the inverse function is:
[tex]\[ g^{-1}(x) = 6x - 1 \][/tex]
