If f:[1, infty) rightarrow[5, infty) is given | vedclass

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(A) Given $f(x) = 3x + \frac{2}{x}$. Let $y = 3x + \frac{2}{x}$.Multiplying by $x$, we get $3x^2 - yx + 2 = 0$.Using the quadratic formula, $x = \frac

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$\frac{1}{6}\left[x+\sqrt{x^2-24}\right]$

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(A) Given $f(x) = 3x + \frac{2}{x}$. Let $y = 3x + \frac{2}{x}$.Multiplying by $x$, we get $3x^2 - yx + 2 = 0$.Using the quadratic formula, $x = \frac{y \pm \sqrt{y^2 - 4(3)(2)}}{2(3)} = \frac{y \pm \sqrt{y^2 - 24}}{6}$.Since the domain is $x \in [1, \infty)$, we must choose the root that satisfies this condition.For $y \geq 5$, $y^2 \geq 25$, so $\sqrt{y^2 - 24} \geq 1$.If we take $x = \frac{y - \sqrt{y^2 - 24}}{6}$, then for $y=5$, $x = \frac{5 - 1}{6} = \frac{2}{3} If we take $x = \frac{y + \sqrt{y^2 - 24}}{6}$, then for $y=5$, $x = \frac{5 + 1}{6} = 1$, which is in the domain.Thus, $f^{-1}(x) = \frac{x + \sqrt{x^2 - 24}}{6}$.

If $f:[1, \infty) \rightarrow[5, \infty)$ is given by $f(x)=3x+\frac{2}{x}$, then $f^{-1}(x)=$

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