Part A.
The given equation is g(x) = 1/3x + 2. First, let g(x) be y so that it would be easy to write in the solution.
y = 1/3x + 2
Next, interchange x and y variables:
x = 1/3y + 2
Solve for y
1/3y = x - 2
y = 3(x-2) = g⁻¹(x) ← This is the inverse function of the original equation
Part B. Graph both equations by assigning arbitrary values of x. You will get their corresponding y values. Plot x against y, then connect the dots. The graph is shown in the attached picture. It is proven that they are inverse functions if when point (a,b) is plotted on g(x), then point (b,a) must also lie on g⁻¹(x). As shown in the picture, P(6,4) in g(x) is the inverse of P(4,6) in g⁻¹(x).
Part C. The proof of inverse functions in functional notation is:
g(g⁻¹(x)) = x and g⁻¹(g(x) = x, for all values of x. Applying this rule,
g(x)=1/3x+2; g⁻¹(x) = 3(x-2)
g(3(x-2) ) = 1/3(3(x-2) )+2 = x - 2 + 2 = x
and
g⁻¹(1/3x+2 ) = 3(1/3x+2-2) = x +6 - 6 = x
Therefore, the equation for g⁻¹(x) is correct.