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

D. [tex]h(x)=3 x-6[/tex]



Answer :

Sure, let's find the inverse of the given function [tex]\( f(x) = \frac{1}{3} x + 2 \)[/tex].

1. Rewrite the function: Start by replacing [tex]\( f(x) \)[/tex] with [tex]\( y \)[/tex]:
[tex]\[ y = \frac{1}{3} x + 2 \][/tex]

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

3. Solve for [tex]\( y \)[/tex]: Isolate [tex]\( y \)[/tex] in the equation.
- First, subtract 2 from both sides to start isolating [tex]\( y \)[/tex]:
[tex]\[ x - 2 = \frac{1}{3} y \][/tex]

- Next, multiply both sides of the equation by 3 to solve for [tex]\( y \)[/tex]:
[tex]\[ 3(x - 2) = y \][/tex]

- Simplify the right hand side:
[tex]\[ y = 3x - 6 \][/tex]

4. Write the inverse function: The function obtained is the inverse of [tex]\( f(x) \)[/tex]. We denote the inverse of [tex]\( f \)[/tex] as [tex]\( h(x) \)[/tex], therefore:
[tex]\[ h(x) = 3x - 6 \][/tex]

So, the inverse of the given function [tex]\( f(x) = \frac{1}{3} x + 2 \)[/tex] is [tex]\( \boxed{h(x) = 3x - 6} \)[/tex].

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