What is the inverse of f(x) 3 - (1/(x+6)^2)
To find the inverse, simply solve for "x" in terms of "y". Replace every "x" in f(x) with 'y' and then set it equal to 'x' and then solve for 'x'
ok so... x = 3 - (1/(y+6)^2) x - 3 = - (1/(y+6)^2 \[\sqrt{x-3} = -1/(y+6)\] And I don't know where to go from there
x - 3 = -1/(y+6)^2 3-x = 1/(y+6)^2 (y+6)^2 = 1/(3-x) now take square root both sides
so the inverse is \[\sqrt{1/(3-x)}\]
taking square root gives us two values for right hand side
x - 3 = -1/(y+6)^2 3-x = 1/(y+6)^2 (y+6)^2 = 1/(3-x) now take square root both sides \((y+6) = \pm \sqrt{\frac{1}{3-x}}\)
subtract 6 both sides to get y
\( (y+6) = \pm \sqrt{\frac{1}{3-x}} \) subtract 6 \( y = -6 \pm \sqrt{\frac{1}{3-x}} \)
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