Question

(a) Find the domain of the function \(f(x) = \frac{{1 - {e^{{x^2}}}}}{{1 - {e^{1 - {x^2}}}}}\).

(b) Find the domain of the function \(f(x) = \frac{{1 + x}}{{{e^m}x}}\).

Short answer

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

Step-by-step solution

Given data

The functions are\(f(x) = \frac{{1 - {e^{{x^2}}}}}{{1 - {e^{1 - {x^2}}}}}\)and \(f(x) = \frac{{1 + x}}{{{e^m}x}}\).

Concept of domain

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

The domain of an exponential functions\(y = {a^x},\;a > 0\)and\(a \ne 1\)is the set of real values.

That is,\( - \infty < x < \infty \).

Find the domain of the function \(f(x) = \frac{{1 - {e^{{x^2}}}}}{{1 - {e^{1 - {x^2}}}}}\)

(a)

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

Since the denominator of \(f(x)\)is \(1 - {e^{1 - {x^2}}}\), the undefined points are obtained as follows:

\(\begin{array}{c}1 - {e^{1 - {x^2}}} = 0\\1 = {e^{1 - {x^2}}}\\{e^0} = {e^{1 - {x^2}}}\end{array}\)

Therefore, \({e^0} = 1\).

Equate the powers as they have the same base and simplify as shown below.

\(\begin{array}{c}0 = 1 - {x^2}\\1 = {x^2}\\x = \pm 1\end{array}\)

Hence, the function is undefined when \(x = - 1\)and \(x = 1\).

Therefore, the domain of the function is \(\{ x \in R\mid x \ne - 1,1\} \) and, the interval notation of the domain of \(f(x)\)is \(( - \infty , - 1) \cup ( - 1,1) \cup (1,\infty )\).

Find the domain of the function \(f(x) = \frac{{1 + x}}{{{e^m}x}}\)

(b)

Consider the denominator of the function \(f(x)\) 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\(( - \infty ,\infty )\).