Let f(x) = a^x (a 0) be written as f(x) = f | vedclass

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(B) We know that any function $f(x)$ can be expressed as the sum of an even function $f_1(x)$ and an odd function $f_2(x)$ as follows:$f_1(x) = \frac{

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$2f_1(x)f_1(y)$

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(B) We know that any function $f(x)$ can be expressed as the sum of an even function $f_1(x)$ and an odd function $f_2(x)$ as follows:$f_1(x) = \frac{f(x) + f(-x)}{2} = \frac{a^x + a^{-x}}{2}$$f_2(x) = \frac{f(x) - f(-x)}{2} = \frac{a^x - a^{-x}}{2}$Now,we calculate $f_1(x + y) + f_1(x - y)$:$f_1(x + y) + f_1(x - y) = \frac{a^{x+y} + a^{-(x+y)}}{2} + \frac{a^{x-y} + a^{-(x-y)}}{2}$$= \frac{1}{2} [a^x a^y + a^{-x} a^{-y} + a^x a^{-y} + a^{-x} a^y]$$= \frac{1}{2} [a^x(a^y + a^{-y}) + a^{-x}(a^y + a^{-y})]$$= \frac{1}{2} (a^x + a^{-x})(a^y + a^{-y})$$= 2 \left( \frac{a^x + a^{-x}}{2} \right) \left( \frac{a^y + a^{-y}}{2} \right)$$= 2 f_1(x) f_1(y)$

Let $f(x) = a^x$ $(a > 0)$ be written as $f(x) = f_1(x) + f_2(x)$,where $f_1(x)$ is an even function and $f_2(x)$ is an odd function. Then $f_1(x + y) + f_1(x - y)$ equals

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