Answer: [tex]f^{-1}(x) = -\sqrt{36-x^2}[/tex]
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Here are the steps to find the inverse
[tex]f(x) = \sqrt{36-x^2}\\\\y = \sqrt{36-x^2}\\\\x = \sqrt{36-y^2} \ \text{ swap x and y; solve for y}\\\\x^2 = 36-y^2\\\\x^2+y^2 = 36\\\\y^2 = 36-x^2\\\\y = \sqrt{36-x^2}\\\\f^{-1}(x) = \sqrt{36-x^2}\\\\[/tex]
Now this would be the answer if our original domain was [tex]0 \le x \le 6[/tex]
However, that's not the case and instead our domain is [tex]-6 \le x \le 0[/tex]
Reflecting any (x,y) point in this domain restriction, over the the line y = x, will have us land on a point with a negative y value. This is because the inverse will swap the x and y values.
Therefore, the inverse function must ultimately be negative.
Currently [tex]f^{-1}(x) = \sqrt{36-x^2}[/tex] is positive since the result of any square root is never negative.
To fix this, we stick a negative out front to get the final answer being [tex]f^{-1}(x) = -\sqrt{36-x^2}[/tex]
