The function f(x) = sec left[ log left( x + s | vedclass

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(A) To determine if the function $f(x) = \sec \left[ \log \left( x + \sqrt{1 + x^2} \right) \right]$ is even or odd,we evaluate $f(-x)$.First,consider

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(A) To determine if the function $f(x) = \sec \left[ \log \left( x + \sqrt{1 + x^2} ight) ight]$ is even or odd,we evaluate $f(-x)$.First,consider the inner function $g(x) = \log \left( x + \sqrt{1 + x^2} ight)$.$g(-x) = \log \left( -x + \sqrt{1 + (-x)^2} ight) = \log \left( \sqrt{1 + x^2} - x ight)$.Multiply and divide by $\left( \sqrt{1 + x^2} + x ight)$:$g(-x) = \log \left( \frac{(\sqrt{1 + x^2} - x)(\sqrt{1 + x^2} + x)}{\sqrt{1 + x^2} + x} ight) = \log \left( \frac{1 + x^2 - x^2}{\sqrt{1 + x^2} + x} ight) = \log \left( \frac{1}{\sqrt{1 + x^2} + x} ight) = -\log \left( x + \sqrt{1 + x^2} ight) = -g(x)$.Since $g(x)$ is an odd function,we have $f(-x) = \sec(g(-x)) = \sec(-g(x))$.Since $\sec(	heta)$ is an even function,$\sec(-g(x)) = \sec(g(x)) = f(x)$.Therefore,$f(-x) = f(x)$,which means the function is an even function.

The function $f(x) = \sec \left[ \log \left( x + \sqrt{1 + x^2} \right) \right]$ is . . . . . . function.

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