Which one of the following yields the inverse for the function $f(x) = 9 - x^2$ when $f$ is restricted to the interval $[0, \infty)$? $\circ$ a. $f^{-1}(x) = \sqrt{9 - x}$ $\circ$ b. $f^{-1}(x) = -\sqrt{9 - x}$ $\circ$ c. $f^{-1}(x) = x$
Added by Alejandro M.
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To do this, we need to switch the roles of x and y and solve for y: x = 9%y2 x/9 = y2 y = ±√(x/9) However, since we are restricted to the interval [0, 100], we only want the positive square root: y = √(x/9) So, the inverse function is f-1(x) = √(x/9). Now, we Show more…
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