To find \((f+g)(x)\), we start by combining the functions \(f(x)\) and \(g(x)\):
- \(f(x) = -2x^2 + 4x - 7\)
- \(g(x) = -\sqrt[3]{x-1} + 2\)
Now we can add them together:
\[ (f+g)(x) = f(x) + g(x) = (-2x^2 + 4x - 7) + (-\sqrt[3]{x-1} + 2) \]
This simplifies to:
\[ (f+g)(x) = -2x^2 + 4x - 7 - \sqrt[3]{x-1} + 2 = -2x^2 + 4x - 5 - \sqrt[3]{x-1} \]
Now regarding the domain:
The cube root function \(-\sqrt[3]{x-1}\) is defined for all real numbers, but there are no restrictions from the quadratic function \(-2x^2 + 4x - 5\) either. Hence, the domain is all real numbers:
So, we have:
\[ (f+g)(x) = -2x^2 + 4x - 5 - \sqrt[3]{x-1} \] and the domain of \((f+g)(x)\) is \((-\infty, \infty)\).
Based on the provided options, the response that matches our calculation is:
(f+g)(x)=−2x^2+4x−x−1−−−−−√3−5 and the domain of (f+g)(x) is (−∞, ∞).