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.

30. \(B\left( u \right) = \sqrt {3u - 2} + \sqrt(3){{2u - 3}}\)

Short answer

The function \(B\left( u \right) = \sqrt {3u - 2} + \sqrt(3){{2u - 3}}\) is continuous at every number in its domain—the domain of \(B\) is \(\left( {\frac{2}{3},\infty } \right)\).

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 the composite 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 function\(B\left( u \right) = \sqrt {3u - 2} + \sqrt(3){{2u - 3}}\)is defined if\(3u - 2 \ge 0\).

\(\begin{array}{c}3u - 2 \ge 0\\3u \ge 2\\u \ge \frac{2}{3}\end{array}\)

It is observed that\(\sqrt(3){{2u - 3}}\)is defined everywhere. Therefore, the domain of\(B\)is\(\left( {\frac{2}{3},\infty } \right)\).

The functions\(\sqrt {3u - 2} \)and\(\sqrt(3){{2u - 3}}\)are continuous on their domain since it is a composite of a root function and a polynomial function.

The function \(B\) is the sum of these two functions. Therefore the function is continuous at everywhere in its domain.

Thus, the function \(B\left( u \right) = \sqrt {3u - 2} + \sqrt(3){{2u - 3}}\) is continuous at every number in its domain.