What is the inverse of [tex]\( f(x) = \frac{1}{3}x + 2 \)[/tex]?

A. [tex]\( h(x) = \frac{1}{3}x + 2 \)[/tex]
B. [tex]\( h(x) = \frac{1}{3}x - 2 \)[/tex]
C. [tex]\( h(x) = 3x - 2 \)[/tex]
D. [tex]\( h(x) = 3x - 6 \)[/tex]



Answer :

To find the inverse of the function [tex]\( f(x) = \frac{1}{3}x + 2 \)[/tex], follow these steps:

1. Rewrite [tex]\( f(x) \)[/tex] as [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1}{3}x + 2 \][/tex]

2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse:
[tex]\[ x = \frac{1}{3}y + 2 \][/tex]

3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x = \frac{1}{3}y + 2 \][/tex]

Subtract 2 from both sides:
[tex]\[ x - 2 = \frac{1}{3}y \][/tex]

To isolate [tex]\( y \)[/tex], multiply both sides of the equation by 3:
[tex]\[ 3(x - 2) = y \][/tex]

4. Simplify the expression:
[tex]\[ y = 3x - 6 \][/tex]

The inverse function [tex]\( h(x) \)[/tex] is therefore:
[tex]\[ h(x) = 3x - 6 \][/tex]

So the correct answer is:
[tex]\[ h(x) = 3x - 6 \][/tex]