explain, using these theorems, why the function is continuous at every number in its domain.\n\n( h ( t ) =…

explain, using these theorems, why the function is continuous at every number in its domain.\n\n( h ( t ) = \frac { cos ( t ^ { 2 } ) } { 1 - e ^ { t } } )\n\n( \bigcirc h ( t ) ) is a rational function, so it is continuous at every number in its domain.\n\n( \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.\n\n( \bigcirc h ( t ) ) is a logarithmic, so it is continuous at every number in its domain.\n\n( \bigcirc h ( t ) ) is a polynomial, so it is continuous at every number in its domain.\n\n( \bigcirc h ( t ) ) is not continuous at every number in its domain.\n\nstate the domain. (enter your answer using interval notation.)

explain, using these theorems, why the function is continuous at every number in its domain.\n\n( h ( t ) = \frac { cos ( t ^ { 2 } ) } { 1 - e ^ { t } } )\n\n( \bigcirc h ( t ) ) is a rational function, so it is continuous at every number in its domain.\n\n( \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.\n\n( \bigcirc h ( t ) ) is a logarithmic, so it is continuous at every number in its domain.\n\n( \bigcirc h ( t ) ) is a polynomial, so it is continuous at every number in its domain.\n\n( \bigcirc h ( t ) ) is not continuous at every number in its domain.\n\nstate the domain. (enter your answer using interval notation.)

Answer

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)\neq0). For (h(t)=\frac{\cos(t^{2})}{1 - e^{t}}), we set (g(t)=1 - e^{t}\neq0). Solving (1 - e^{t}\neq0), we get (e^{t}\neq1). 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))