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What is the inverse of the function [tex]f(x) = \frac{1}{9} x + 2[/tex]?
A. [tex]h(x) = 18x - 2[/tex] B. [tex]h(x) = 9x - 18[/tex] C. [tex]h(x) = 9x + 18[/tex] D. [tex]h(x) = 18x + 2[/tex]
To find the inverse of the function [tex]\( f(x) = \frac{1}{9}x + 2 \)[/tex]:
1. Start with the function and let it equal [tex]\( y \)[/tex]: [tex]\[
y = \frac{1}{9}x + 2
\][/tex]
2. Swap [tex]\( x \)[/tex] and [tex]\( y \)[/tex] to find the inverse function: [tex]\[
x = \frac{1}{9}y + 2
\][/tex]
3. Solve for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex]: - First, isolate the term containing [tex]\( y \)[/tex]: [tex]\[
x - 2 = \frac{1}{9}y
\][/tex] - Next, multiply both sides by 9 to solve for [tex]\( y \)[/tex]: [tex]\[
9(x - 2) = y
\][/tex] - Simplify the right-hand side: [tex]\[
y = 9x - 18
\][/tex]
4. Write the inverse function [tex]\( h(x) \)[/tex]: [tex]\[
h(x) = 9x - 18
\][/tex]
Therefore, the inverse of the function [tex]\( f(x) = \frac{1}{9}x + 2 \)[/tex] is: [tex]\[
h(x) = 9x - 18
\][/tex]
Hence, the correct answer is [tex]\( h(x) = 9x - 18 \)[/tex].
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