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

Let the function $f:[0,1] \rightarrow \mathbb{R}$ be defined by $f(x) = \frac{4^x}{4^x+2}$. Then the value of $f\left(\frac{1}{40}\right) + f\left(\frac{2}{40}\right) + f\left(\frac{3}{40}\right) + \dots + f\left(\frac{39}{40}\right) - f\left(\frac{1}{2}\right)$ is:

If $f(x) = 4x^3 + 3x^2 + 3x + 4$,then $x^3 f\left( \frac{1}{x} \right)$ is

Let $f(x) = \frac{2^{x+2} + 16}{2^{2x+1} + 2^{x+4} + 32}$. Then the value of $8 \left( f \left( \frac{1}{15} \right) + f \left( \frac{2}{15} \right) + \dots + f \left( \frac{59}{15} \right) \right)$ is equal to

Let $f(x)$ and $g(x)$ be two continuous functions defined from $R \rightarrow R$,such that $f(x_1) > f(x_2)$ and $g(x_1) < g(x_2)$ for all $x_1 > x_2$. Then the solution set of $f(g(\alpha^2 - 2\alpha)) > f(g(3\alpha - 4))$ is

If $f(x) = \frac{1}{\sqrt{x + 2\sqrt{2x - 4}}} + \frac{1}{\sqrt{x - 2\sqrt{2x - 4}}}$ for $x > 2$,then $f(11) = $