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

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

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An odd function

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(B) To determine if the function $f(x) = \log (x + \sqrt {{x^2} + 1} )$ is even or odd,we evaluate $f(-x)$.$f(-x) = \log (-x + \sqrt {(-x)^2 + 1} )$$f(-x) = \log (-x + \sqrt {x^2 + 1} )$We can rewrite the argument of the logarithm by multiplying and dividing by the conjugate $(\sqrt {x^2 + 1} + x)$:$f(-x) = \log \left( \frac{(\sqrt {x^2 + 1} - x)(\sqrt {x^2 + 1} + x)}{\sqrt {x^2 + 1} + x} \right)$$f(-x) = \log \left( \frac{(x^2 + 1) - x^2}{\sqrt {x^2 + 1} + x} \right)$$f(-x) = \log \left( \frac{1}{\sqrt {x^2 + 1} + x} \right)$Using the property $\log(1/a) = -\log(a)$:$f(-x) = -\log (x + \sqrt {x^2 + 1} )$$f(-x) = -f(x)$Since $f(-x) = -f(x)$,the function $f(x)$ is an odd function.

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

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