what is the inverse of the function $f(x) = \\frac{1}{9}x + 2$?\\n\\n$\\circ$ $h(x) = 18x - 2$\\n$\\circ$…

what is the inverse of the function $f(x) = \\frac{1}{9}x + 2$?\\n\\n$\\circ$ $h(x) = 18x - 2$\\n$\\circ$ $h(x) = 9x - 18$\\n$\\circ$ $h(x) = 9x + 18$\\n$\\circ$ $h(x) = 18x + 2$

what is the inverse of the function $f(x) = \\frac{1}{9}x + 2$?\\n\\n$\\circ$ $h(x) = 18x - 2$\\n$\\circ$ $h(x) = 9x - 18$\\n$\\circ$ $h(x) = 9x + 18$\\n$\\circ$ $h(x) = 18x + 2$

Answer

Answer:

B. h(x) = 9x - 18

Explanation:

Step1: Replace f(x) with y

$y = \frac{1}{9}x + 2$

Step2: Swap x and y

$x = \frac{1}{9}y + 2$

Step3: Solve for y

Subtract 2: $x - 2 = \frac{1}{9}y$
Multiply by 9: $y = 9(x - 2) = 9x - 18$