The function f(x) = log (x + sqrt{x^2 + 1}) i | vedclass

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(A) To check if the function is even or odd, we evaluate $f(-x)$.$f(-x) = \log (-x + \sqrt{(-x)^2 + 1}) = \log (-x + \sqrt{x^2 + 1})$.Now, multiply an

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An odd function

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(A) To check if the function is even or odd, we evaluate $f(-x)$.$f(-x) = \log (-x + \sqrt{(-x)^2 + 1}) = \log (-x + \sqrt{x^2 + 1})$.Now, multiply and divide by $(x + \sqrt{x^2 + 1})$ inside the logarithm:$f(-x) = \log \left( \frac{(\sqrt{x^2 + 1} - x)(\sqrt{x^2 + 1} + x)}{\sqrt{x^2 + 1} + x} \right)$.Using the identity $(a-b)(a+b) = a^2 - b^2$:$f(-x) = \log \left( \frac{(x^2 + 1) - x^2}{\sqrt{x^2 + 1} + x} \right) = \log \left( \frac{1}{x + \sqrt{x^2 + 1}} \right)$.Using the property $\log(1/a) = -\log(a)$:$f(-x) = -\log(x + \sqrt{x^2 + 1}) = -f(x)$.Since $f(-x) = -f(x)$, the function is an odd function.

The function $f(x) = \log (x + \sqrt{x^2 + 1})$ is

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