To verify that [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex] are inverses of each other, we need to confirm that composing the functions in both orders yields the identity function [tex]\( x \)[/tex]. That means we need to show that [tex]\( f(g(x)) = x \)[/tex] and [tex]\( g(f(x)) = x \)[/tex].
### Step 1: Check to see if [tex]\( f(g(x)) = x \)[/tex].
Given [tex]\( f(x) = 3x - 12 \)[/tex] and [tex]\( g(x) = \frac{1}{3}x + 4 \)[/tex], we first compose [tex]\( f \)[/tex] and [tex]\( g \)[/tex]:
[tex]\[
f(g(x)) = f\left(\frac{1}{3}x + 4\right)
\][/tex]
Now, we substitute [tex]\( \frac{1}{3}x + 4 \)[/tex] into the function [tex]\( f \)[/tex]:
[tex]\[
f\left(\frac{1}{3}x + 4\right) = 3\left(\frac{1}{3}x + 4\right) - 12
\][/tex]
Next, we distribute the 3:
[tex]\[
3\left(\frac{1}{3}x + 4\right) = 3 \cdot \frac{1}{3}x + 3 \cdot 4 = x + 12
\][/tex]
Then we subtract 12:
[tex]\[
x + 12 - 12 = x
\][/tex]
Thus,
[tex]\[
f(g(x)) = x
\][/tex]
### Step 2: Check to see if [tex]\( g(f(x)) = x \)[/tex].
Now we compose [tex]\( g \)[/tex] and [tex]\( f \)[/tex]:
[tex]\[
g(f(x)) = g(3x - 12)
\][/tex]
We substitute [tex]\( 3x - 12 \)[/tex] into the function [tex]\( g \)[/tex]:
[tex]\[
g(3x - 12) = \frac{1}{3}(3x - 12) + 4
\][/tex]
Next, we distribute the [tex]\( \frac{1}{3} \)[/tex]:
[tex]\[
\frac{1}{3}(3x - 12) = \frac{1}{3} \cdot 3x - \frac{1}{3} \cdot 12 = x - 4
\][/tex]
Then we add 4:
[tex]\[
x - 4 + 4 = x
\][/tex]
So,
[tex]\[
g(f(x)) = x
\][/tex]
Having confirmed both compositions result in the identity function, we can conclude that [tex]\( f(x) = 3x - 12 \)[/tex] and [tex]\( g(x) = \frac{1}{3}x + 4 \)[/tex] are indeed inverses of each other.
