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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)=3(x-2)[/tex]

Sagot :

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

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

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

3. Solve for [tex]\( y \)[/tex]:
[tex]\[ x = \frac{1}{3} y + 2 \][/tex]
Multiply both sides by 3 to clear the fraction:
[tex]\[ 3x = y + 6 \][/tex]
Subtract 6 from both sides to isolate [tex]\( y \)[/tex]:
[tex]\[ y = 3x - 6 \][/tex]

4. Thus, the inverse function is:
[tex]\[ h(x) = 3x - 6 \][/tex]

So, the correct inverse function is [tex]\( h(x) = 3x - 6 \)[/tex].
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