If the function f : R to R is defined by f(x) | vedclass

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(A) Given $f(x) = \log_a(x + \sqrt{x^2 + 1})$.Let $y = f(x)$,so $y = \log_a(x + \sqrt{x^2 + 1})$.By the definition of logarithm,$a^y = x + \sqrt{x^2 +

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$\frac{a^x - a^{-x}}{2}$

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(A) Given $f(x) = \log_a(x + \sqrt{x^2 + 1})$.Let $y = f(x)$,so $y = \log_a(x + \sqrt{x^2 + 1})$.By the definition of logarithm,$a^y = x + \sqrt{x^2 + 1}$.Now,consider $a^{-y} = \frac{1}{a^y} = \frac{1}{x + \sqrt{x^2 + 1}}$.Rationalizing the denominator: $a^{-y} = \frac{\sqrt{x^2 + 1} - x}{(\sqrt{x^2 + 1} + x)(\sqrt{x^2 + 1} - x)} = \frac{\sqrt{x^2 + 1} - x}{(x^2 + 1) - x^2} = \sqrt{x^2 + 1} - x$.We have two equations:$1) \sqrt{x^2 + 1} + x = a^y$$2) \sqrt{x^2 + 1} - x = a^{-y}$Subtracting equation $(2)$ from equation $(1)$:$(\sqrt{x^2 + 1} + x) - (\sqrt{x^2 + 1} - x) = a^y - a^{-y}$$2x = a^y - a^{-y}$$x = \frac{a^y - a^{-y}}{2}$.Since $x = f^{-1}(y)$,we have $f^{-1}(y) = \frac{a^y - a^{-y}}{2}$.Replacing $y$ with $x$,we get $f^{-1}(x) = \frac{a^x - a^{-x}}{2}$.

If the function $f : R \to R$ is defined by $f(x) = \log_a(x + \sqrt{x^2 + 1})$,where $(a > 0, a \neq 1)$,then $f^{-1}(x)$ is

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