Question

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

f(t) = e^{-t^{2}}\ln(7 + t^{2})

\\( f(t) \\) is a root function, so it is continuous at every number in its domain.
\\( f(t) \\) is the product of functions that are continuous on the domain of \\( f(t) \\), so it is continuous at every number in its domain.
\\( f(t) \\) is a rational function, so it is continuous at every number in its domain.
\\( f(t) \\) is a polynomial function, so it is continuous at every number in its domain.
\\( f(t) \\) is not continuous at every number in its domain.

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

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Explanation:

Step1: Analyze the components of \( f(t) \)

The function \( f(t)=e^{-t^{2}}\ln(7 + t^{2}) \) is a product of two functions \( g(t)=e^{-t^{2}} \) and \( h(t)=\ln(7 + t^{2}) \).

• The exponential function \( y = e^{u}\) with \( u=-t^{2}\) is continuous for all real \( t\) since the exponential function \( y = e^{x}\) is continuous for all \( x\in R\) and the polynomial \( u=-t^{2}\) is continuous for all \( t\in R\). By the composition - of - continuous - functions theorem (if \( u = g(x)\) is continuous at \( x = a\) and \( y = f(u)\) is continuous at \( u = g(a)\), then \( y=(f\circ g)(x)\) is continuous at \( x = a\)), \( g(t)=e^{-t^{2}}\) is continuous for all \( t\in R\).
• For the logarithmic function \( y=\ln(u)\), the argument \( u = 7 + t^{2}\). Since \(t^{2}\geq0\) for all \(t\in R\), then \(u=7 + t^{2}\geq7>0\) for all \(t\in R\). The function \(u = 7 + t^{2}\) is a polynomial (continuous for all \(t\in R\)) and the logarithmic function \(y=\ln(u)\) is continuous for \(u>0\). By the composition - of - continuous - functions theorem, \(h(t)=\ln(7 + t^{2})\) is continuous for all \(t\in R\).

Step2: Use the product - of - continuous - functions theorem

By the product - of - continuous - functions theorem (if \(f(x)\) and \(g(x)\) are continuous at \(x = a\), then \(h(x)=f(x)g(x)\) is continuous at \(x = a\)), since \(g(t)=e^{-t^{2}}\) and \(h(t)=\ln(7 + t^{2})\) are continuous for all \(t\in R\), the function \(f(t)=e^{-t^{2}}\ln(7 + t^{2})\) is continuous for all \(t\in R\) (its domain).

Step3: Find the domain of \(f(t)\)

For the function \(y = f(t)=e^{-t^{2}}\ln(7 + t^{2})\), the domain of \(e^{-t^{2}}\) is \(R\) (all real numbers) and for \(y=\ln(7 + t^{2})\), since \(7 + t^{2}>0\) for all \(t\in R\) (because \(t^{2}\geq0\) for all \(t\in R\) and \(7>0\)), the domain of \(f(t)\) is the set of all real numbers. In interval notation, the domain is \((-\infty,\infty)\)

Answer:

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