# Chapter 1 - Section 1.5 - Exponential Functions - Exercises - Page 38: 21

The domain of $f(x)$ is $(-\infty,\infty)$ The range of $f(x)$ is $(0,\frac{1}{2})$.

#### Work Step by Step

$$f(x)=\frac{1}{2+e^x}$$ 1) Domain: $f(x)$ is defined on $(-\infty,\infty)$ except where $2+e^x=0$ However, since $e^x\gt0$ for all $x\in R$, $2+e^x\gt0$ for all $x\in R$. Therefore, the domain of $f(x)$ is $(-\infty,\infty)$ 2) Range: As stated above, $e^x\gt0$ for all $x\in(-\infty,\infty)$ So $2+e^{x}\gt2$ for all $x\in(-\infty,\infty)$ That means $\frac{1}{2+e^x}\lt\frac{1}{2}$ for $x\in(-\infty,\infty)$ Also, since both $2+e^x\gt0$ and $1\gt0$, we have $\frac{1}{2+e^x}\gt0$ Therefore, overall, the range of $f(x)$ is $(0,\frac{1}{2})$.

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