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

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(C) Given $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$.To find $f^{-1}(3)$, we set $f(x) = y$, which implies $x = f^{-1}(y)$.$\frac{1+\sqrt{1+4 \log_2 x}}

Vedclass · Accepted Answer

$64$

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(C) Given $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$.To find $f^{-1}(3)$, we set $f(x) = y$, which implies $x = f^{-1}(y)$.$\frac{1+\sqrt{1+4 \log_2 x}}{2} = y$$\sqrt{1+4 \log_2 x} = 2y - 1$Squaring both sides:$1 + 4 \log_2 x = (2y - 1)^2$$1 + 4 \log_2 x = 4y^2 - 4y + 1$$4 \log_2 x = 4y^2 - 4y$$\log_2 x = y^2 - y$$x = 2^{y^2 - y}$Thus, $f^{-1}(y) = 2^{y^2 - y}$.Now, substitute $y = 3$:$f^{-1}(3) = 2^{3^2 - 3}$$f^{-1}(3) = 2^{9 - 3}$$f^{-1}(3) = 2^6 = 64$.

If $f:[1, \infty) \rightarrow [1, \infty)$ is defined by $f(x) = \frac{1+\sqrt{1+4 \log_2 x}}{2}$, then $f^{-1}(3) =$

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