If the function f:[1, infty) to [1, infty) is | vedclass

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(B) Given $f(x) = 2^{x(x - 1)}$.To find the inverse,let $y = f(x) = 2^{x(x - 1)}$.Taking $\log_2$ on both sides,we get $\log_2 y = x(x - 1)$.This simp

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$\frac{1}{2}(1 + \sqrt{1 + 4\log_2 x})$

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(B) Given $f(x) = 2^{x(x - 1)}$.To find the inverse,let $y = f(x) = 2^{x(x - 1)}$.Taking $\log_2$ on both sides,we get $\log_2 y = x(x - 1)$.This simplifies to the quadratic equation $x^2 - x - \log_2 y = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$,we get $x = \frac{1 \pm \sqrt{1 + 4\log_2 y}}{2}$.Since the domain of $f$ is $[1, \infty)$,we must have $x \ge 1$.If we choose the negative sign,$x = \frac{1 - \sqrt{1 + 4\log_2 y}}{2} Thus,we must choose the positive sign: $x = \frac{1 + \sqrt{1 + 4\log_2 y}}{2}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{1}{2}(1 + \sqrt{1 + 4\log_2 x})$.

If the function $f:[1, \infty) \to [1, \infty)$ is defined by $f(x) = 2^{x(x - 1)}$,then $f^{-1}(x)$ is

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