Question 1196463
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The standard template or outline to finding the inverse is this:
1) Replace f(x) with y
2) Swap x and y
3) Solve for y to find *[tex \Large \text{f}^{-1}(\text{x})]


So the steps look like this:
{{{f(x) = 6 - x^2}}}


{{{y = 6 - x^2}}} Apply step 1


{{{x = 6 - y^2}}} Apply step 2


{{{x+y^2 = 6}}}


{{{y^2 = 6-x}}}


{{{y = sqrt(6-x)}}}


The domain of f(x) is {{{x >= 0}}}
The domain and range swap when going from the original function to the inverse (since the x and y swap roles)
This means the range of the inverse is {{{y >= 0}}} and it further means we'll focus on the positive version of the square root only. We won't have "plus/minus".


Answer: The inverse is *[tex \Large \text{f}^{-1}(\text{x}) = \sqrt{6-\text{x}}]
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