Question

at what points is the function ( y = (9x - 8)^{\frac{1}{3}} ) continuous? describe the set of x-values where the function is continuous using interval notation (left \frac{8}{9}, infty
ight)) (simplify your answer.)

Explanation:

Step1: Rewrite the function

Recall that \(a^{\frac{1}{n}}=\sqrt[n]{a}\). So, \(y=(9x - 8)^{\frac{1}{3}}=\sqrt[3]{9x - 8}\).

Step2: Analyze the domain of the cube - root function

The cube - root function \(f(t)=\sqrt[3]{t}\) is defined for all real numbers \(t\) (because for any real number \(t\), we can find a real number \(s\) such that \(s^{3}=t\), including when \(t\) is negative, zero, or positive). In our function, \(t = 9x-8\). Since \(9x - 8\) is a linear function (a polynomial of degree 1) and polynomials are defined for all real numbers \(x\), and the cube - root function is defined for all real values of its argument, the composition of the cube - root function and the linear function \(9x - 8\) is defined for all real numbers \(x\).

Wait, there is a mistake in the initial given answer. Let's re - examine:

The function \(y=(9x - 8)^{\frac{1}{3}}=\sqrt[3]{9x - 8}\). The cube root function \(\sqrt[3]{u}\) is continuous for all real numbers \(u\) (because the derivative of \(y = \sqrt[3]{u}=u^{\frac{1}{3}}\) is \(y^\prime=\frac{1}{3}u^{-\frac{2}{3}}=\frac{1}{3u^{\frac{2}{3}}}\), which is defined for all \(u
eq0\), and the function is continuous at \(u = 0\) as well, since \(\lim_{u
ightarrow0}\sqrt[3]{u}=0=\sqrt[3]{0}\)). And \(u = 9x - 8\) is a linear function, which is continuous for all real \(x\). By the theorem of continuity of composite functions: if \(f(u)\) is continuous at \(u = a\) and \(u = g(x)\) is continuous at \(x = c\) with \(g(c)=a\), then \(f(g(x))\) is continuous at \(x = c\). Since \(f(u)=\sqrt[3]{u}\) is continuous for all \(u\in\mathbb{R}\) and \(g(x)=9x - 8\) is continuous for all \(x\in\mathbb{R}\), the composite function \(y = f(g(x))=(9x - 8)^{\frac{1}{3}}\) is continuous for all real numbers \(x\).

So the correct interval notation for the set of \(x\) - values where the function is continuous is \((-\infty,\infty)\).

Answer:

\((-\infty,\infty)\)