The function f(x) = frac{e^{|x|} - e^{-x}}{e^ | vedclass

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(D) Let $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}\right)$.For $x \ge 0$,$|x| = x$,so $f(x) = \frac{e^x - e^{-x}}{e^x + e

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Not bijective

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(D) Let $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}ight)$.For $x \ge 0$,$|x| = x$,so $f(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}ight) = 	anh(x) + \cos^3\left(\frac{x}{2}ight)$.For $x Since $f(x)$ is not symmetric about the $y$-axis $(f(x) 
eq f(-x))$,it is not an even function.Since $f(x)$ is not symmetric about the origin $(f(-x) 
eq -f(x))$,it is not an odd function.As $x 	o \infty$,$f(x) 	o 1 + 0 = 1$. As $x 	o -\infty$,$f(x) = \cos^3(x/2)$,which oscillates between $-1$ and $1$.Since the function is not monotonic and its range is not $R$,it is not surjective.Since it is not surjective,it cannot be bijective. Thus,the function is not bijective.

The function $f(x) = \frac{e^{|x|} - e^{-x}}{e^x + e^{-x}} + \cos^3\left(\frac{x}{2}\right)$ from $R$ to itself is

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