Let f: [4, infty) to [1, infty) be a function | vedclass

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(B) Given $f(x) = 5^{x(x - 4)} = y$. Taking $\log_5$ on both sides,we get $\log_5 y = x(x - 4)$. $\log_5 y = x^2 - 4x$. To solve for $x$,we complete t

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$2 + \sqrt{4 + \log_5 x}$

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(B) Given $f(x) = 5^{x(x - 4)} = y$. Taking $\log_5$ on both sides,we get $\log_5 y = x(x - 4)$. $\log_5 y = x^2 - 4x$. To solve for $x$,we complete the square: $x^2 - 4x + 4 = \log_5 y + 4$. $(x - 2)^2 = \log_5 y + 4$. $x - 2 = \pm \sqrt{\log_5 y + 4}$. Since the domain is $x \in [4, \infty)$,we must have $x - 2 \ge 2$,so we take the positive root: $x = 2 + \sqrt{\log_5 y + 4}$. Replacing $y$ with $x$,we get $f^{-1}(x) = 2 + \sqrt{4 + \log_5 x}$.

Let $f: [4, \infty) \to [1, \infty)$ be a function defined by $f(x) = 5^{x(x - 4)}$. Then $f^{-1}(x)$ is:

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