To find \((f+g+h)(x)\), we need to add the three functions together:
- \(f(x) = x^5 - 2x^4 + 5x - 11\)
- \(g(x) = 9x - 1\)
- \(h(x) = \sqrt[3]{x} + 6\)
Let's add them step by step.
\[ (f + g + h)(x) = f(x) + g(x) + h(x) \]
Substituting the functions in:
\[ f(x) + g(x) + h(x) = (x^5 - 2x^4 + 5x - 11) + (9x - 1) + (\sqrt[3]{x} + 6) \]
Now, combine like terms:
- The \(x^5\) term: \(x^5\)
- The \(x^4\) term: \(-2x^4\)
- The \(x\) terms: \(5x + 9x = 14x\)
- The constant terms: \(-11 - 1 + 6 = -6\)
- The cube root term: \(\sqrt[3]{x}\)
So, we get:
\[ (f + g + h)(x) = x^5 - 2x^4 + 14x + \sqrt[3]{x} - 6 \]
Next, we determine the domain of \((f + g + h)(x)\). The function \(g(x) = 9x - 1\) and \(f(x) = x^5 - 2x^4 + 5x - 11\) are polynomials and are defined for all real \(x\). The term \(h(x) = \sqrt[3]{x} + 6\) is also defined for all \(x\) since the cube root function is defined for all real numbers.
Therefore, the combined domain of \( (f+g+h)(x) \) is:
\[ \text{Domain: } (-\infty, \infty) \]
Putting it all together, the answer is:
\[ (f+g+h)(x) = x^5 - 2x^4 + 14x + \sqrt[3]{x} - 6 \quad \text{and the domain is } (-\infty, \infty). \]