To find \((f + g + h)(x)\), we need to add the functions \(f(x)\), \(g(x)\), and \(h(x)\).
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Function Definitions:
- \(f(x) = x^5 - 2x^4 + 5x - 11\)
- \(g(x) = 9x - 1\)
- \(h(x) = \sqrt[3]{x} + 6\)
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Combining the Functions: \[ (f + g + h)(x) = f(x) + g(x) + h(x) \] \[ = (x^5 - 2x^4 + 5x - 11) + (9x - 1) + (\sqrt[3]{x} + 6) \]
Now, combining like terms:
- The \(x^5\) term: \(x^5\)
- The \(x^4\) term: \(-2x^4\)
- The \(x\) terms: \(5x + 9x = 14x\)
- The constant terms: \(-11 - 1 + 6 = -6\)
Therefore, \[ (f + g + h)(x) = x^5 - 2x^4 + 14x - 6 + \sqrt[3]{x} \]
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Domain of \((f + g + h)(x)\):
- The function \(f(x)\) is a polynomial, which is defined for all \(x\).
- The function \(g(x)\) is also a polynomial, defined for all \(x\).
- The function \(h(x) = \sqrt[3]{x} + 6\) is defined for all \(x\) since the cube root function is defined for all real numbers.
Since all individual functions are defined for all real numbers, the combined function \((f + g + h)(x)\) also has a domain of: \[ (-\infty, \infty) \]
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Final Results: \[ (f + g + h)(x) = x^5 - 2x^4 + 14x - 6 + \sqrt[3]{x} \] The domain of \((f + g + h)(x)\) is: \[ (-\infty, \infty) \]
Thus, the correct choice would be: \((f + g + h)(x) = x^5 - 2x^4 + 14x - 6 + \sqrt[3]{x}\) and the domain of \((f + g + h)(x)\) is \((-\infty, \infty)\).