Question

Find the domain of each function.

17. (a) \(f\left( x \right) = \frac{{{\bf{1}} - {e^{{x^{\bf{2}}}}}}}{{{\bf{1}} - {e^{{\bf{1}} - {x^{\bf{2}}}}}}}\)

(b) \(f\left( x \right) = \frac{{{\bf{1}} + x}}{{{e^{cosx}}}}\)

Short answer

(a) The domain of the function is \(\left\{ {\left. x \right|x \ne \pm 1} \right\}\), or \(\left( { - \infty , - 1} \right) \cup \left( { - 1,1} \right) \cup \left( {1,\infty } \right)\).

(b) The domain of the function is \(\mathbb{R}\).

Step-by-step solution

The domain of the function

The function \(f\left( x \right) = \frac{{1 - {e^{{x^2}}}}}{{1 - {e^{1 - {x^2}}}}}\) is defined if the denominator of the rational function is not 0.

So,\(1 - {e^{1 - {x^2}}} \ne 0\).

Now, the value of\(x\)for which the denominator is 0 is shown below:

\(\begin{aligned}1 - {e^{1 - {x^2}}} &= 0\\{e^{1 - {x^2}}} &= 1\\{e^{1 - {x^2}}} &= {e^0}\end{aligned}\)

Solve further,

\(\begin{aligned}1 - {x^2} &= 0\\{x^2} &= 1\\x &= \pm 1\end{aligned}\)

So, the function is defined when\(x \ne \pm 1\).

Thus, the domain of the given function is \(\left\{ {\left. x \right|x \ne \pm 1} \right\}\), or \(\left( { - \infty , - 1} \right) \cup \left( { - 1,1} \right) \cup \left( {1,\infty } \right)\).

The domain of the function

The function \(f\left( x \right) = \frac{{1 + x}}{{{e^{\cos x}}}}\) is defined if the denominator of the rational function is not 0.

So,\({e^{\cos x}} \ne 0\).

Here, there is no value of\(x\)for which the denominator is 0. So,\(x \in \mathbb{R}\).

Thus, the domain of the function is \(\mathbb{R}\).