Question

explain, using these theorems, why the function is continuous at every number in its domain.

( h ( t ) = \frac { cos ( t ^ { 2 } ) } { 1 - e ^ { t } } )

( \bigcirc h ( t ) ) is a rational function, so it is continuous at every number in its domain.

( \bigcirc h ( t ) ) is the quotient of functions that are continuous on the domain of ( h ( t ) ), so it is continuous at every number in its domain.

( \bigcirc h ( t ) ) is a logarithmic, so it is continuous at every number in its domain.

( \bigcirc h ( t ) ) is a polynomial, so it is continuous at every number in its domain.

( \bigcirc h ( t ) ) is not continuous at every number in its domain.

state the domain. (enter your answer using interval notation.)

Explanation:

Step1: Analyze the function type

The function \(h(t)=\frac{\cos(t^{2})}{1 - e^{t}}\) is a quotient of two functions \(f(t)=\cos(t^{2})\) and \(g(t)=1 - e^{t}\). The cosine function \(y = \cos(u)\) is continuous for all \(u\in R\), and \(u = t^{2}\) (a polynomial, which is continuous for all \(t\in R\)). By the composition of continuous functions theorem, \(\cos(t^{2})\) is continuous for all \(t\in R\). The exponential function \(y = e^{t}\) is continuous for all \(t\in R\), and \(1 - e^{t}\) (a linear combination of continuous functions) is continuous for all \(t\in R\).

Step2: Apply the quotient of continuous functions theorem

The quotient of two continuous functions \(y=\frac{f(t)}{g(t)}\) is continuous at all points \(t\) where \(g(t)
eq0\). For \(h(t)=\frac{\cos(t^{2})}{1 - e^{t}}\), we set \(g(t)=1 - e^{t}
eq0\). Solving \(1 - e^{t}
eq0\), we get \(e^{t}
eq1\). Since \(e^{t}=1\) when \(t = 0\) (because \(y = e^{t}\) and \(e^{0}=1\)), the domain of \(h(t)\) is all real numbers except \(t = 0\).

Answer:

• The correct option is: \(h(t)\) is the quotient of functions that are continuous on the domain of \(h(t)\), so it is continuous at every number in its domain.
• The domain of \(h(t)\) is \((-\infty,0)\cup(0,\infty)\)