Chapter 1.5, Problem 19E

(a)

To determine

To find: The domain of the function $f\left(x\right)=\frac{1-e^{x^2}}{1-e^{1-x^2}}$.

Answer to Problem 19E

The domain of the function f (x) is $\left(-\infty ,-1\right)\cup \left(-1,1\right)\cup \left(1,\infty \right)$.

Explanation of Solution

Definition used:

The domain is the set of all input values of the function for which the function is real and defined.

Calculation:

Consider the denominator of the function $f\left(x\right)$ and equate to zero to obtain the undefined points.

Since the denominator of $f\left(x\right)$ is $1-e^{1-x^2}$, the undefined points are obtained as shown below.

$\begin{array}{l}1-e^{1-x^2}=0\\ 1=e^{1-x^2}\\ e^0=e^{1-x^2}\left[\because e^0=1\right]\end{array}$

Equate the powers as they have the same base and simplify as follows.

$\begin{array}{l}0=1-x^2\\ 1=x^2\\ x=\pm 1\end{array}$

Hence, the function is undefined when $x=-1$ and $x=1$.

Therefore, the domain of the function is $\left\{x\in ℝ|x\ne -1,1\right\}$.

And, the interval notation of the domain of $f\left(x\right)$ is $\left(-\infty ,-1\right)\cup \left(-1,1\right)\cup \left(1,\infty \right)$.

(b)

To determine

To find: The domain of the function $f\left(x\right)=\frac{1+x}{e^{\cos x}}$.

Answer to Problem 19E

The domain of the function f (x) is $\left(-\infty ,\infty \right)$.

Explanation of Solution

Consider the denominator of the function $f\left(x\right)$ and equate to zero to obtain the undefined points, $e^{\cos x}=0$.

Since the output of the exponential function can never be 0, the function is defined for any values of x.

Therefore, the domain of the given function is the set of all real numbers, which is $\left(-\infty ,\infty \right)$.