Question

27-34: Explain using theorem 4,5,7, and 9, why the function is continuous at every number in its domain. State the domain.

32.\(f\left( t \right) = {e^{ - {t^2}}}\ln \left( {1 + {t^2}} \right)\)

Short answer

The function \(f\left( t \right)\) is continuous everywhere in its domain—the domain of \(f\left( t \right)\) is \(\left( { - \infty ,\infty } \right)\).

Step-by-step solution

Theorems of continuity

When the functions say \(f\) and \(g\) arecontinuous at some number. Then, the following functions are also continuous:

1. \(f + g\)
2. \(f - g\)
3. \(cf\)
4. \(fg\)
5. \(\frac{f}{g},\,\,{\mathop{\rm if}\nolimits} \,\,g\left( a \right) \ne 0\)

Eachpolynomial iscontinuous on \(\mathbb{R} = \left( { - \infty ,\infty } \right)\). Anyrational functionis continuous in its domain.

When \(f\), and \(g\) are continuous at some number, then thecomposite function\(f \circ g\)is \(\left( {f \circ g} \right)\left( x \right) = f\left( {g\left( x \right)} \right)\) also continuous.

Explain why the function is continuous at every number

The domain of the function \(f\left( t \right)\) is \(\left( { - \infty ,\infty } \right)\) because \(1 + {t^2} > 0\).

The function \({e^{ - {t^2}}}\) and the composite function \(\ln \left( {1 + {t^2}} \right)\) are continuous everywhere on their domain.

Therefore, the function \(f\left( t \right)\) is continuous everywhere in its domain.

Thus, the function \(f\left( t \right)\) is continuous everywhere in its domain.