Question

(a) Find the domain of the function \(g(t) = \sin \;\left( {{e^x} - 1} \right)\).

(b) Find the domain of the function \(g(t) = \sqrt {1 - {2^t}} \).

Short answer

(a) The domain of the function \(g(t)\) is \(( - \infty ,\infty )\).

(b) The domain of the function \(g(t)\) is \((\infty ,0)\).

Step-by-step solution

Given data

The functions are\(g(t) = \sin \;\left( {{e^x} - 1} \right)\) and \(g(t) = \sqrt {1 - {2^t}} \).

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 \(g(t) = \sin \;\left( {{e^x} - 1} \right)\)

(a)

Notice that the sine function is defined on all real numbers and the exponential function also is defined for all real numbers.

Thus, the function \(g(t) = \sin \left( {{e^t} - 1} \right)\) is defined on set of all real numbers.

Therefore, the domain of the function is whole \(R\).

Thus, the interval notation of the domain of the function is \(( - \infty ,\infty )\).

Find the domain of the function \(g(t) = \sqrt {1 - {2^t}} \)

(b)

Since, the radicand cannot take negative values, \(1 - {2^t} \ge 0\).

Thus, \(1 \ge {2^t}\).

Express 1 as a power of 2 and obtain the value of \(t\) as follows:

\(\begin{array}{c}{2^0} \ge {2^t}\\0 \ge t\\t \le 0\end{array}\)

Thus, the function \(g(t) = \sqrt {1 - {2^t}} \) is defined when \(t \le 0\).

Therefore, the domain of the function is \(t \le 0\).

Thus, the interval notation of the domain of the function is \((\infty ,0)\).