The domain of the real-valued function f(x) = | vedclass

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(C) Given function: $f(x) = \sqrt[3]{\frac{x-2}{2x^2-7x+5}} + \log(x^2-x-2)$.For the domain,the cube root is defined for all real values where the den

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$(-\infty, -1) \cup (2, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$

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(C) Given function: $f(x) = \sqrt[3]{\frac{x-2}{2x^2-7x+5}} + \log(x^2-x-2)$.For the domain,the cube root is defined for all real values where the denominator is non-zero,and the logarithm is defined for positive arguments.$1$. For the denominator of the cube root: $2x^2-7x+5 
eq 0$ $\Rightarrow (2x-5)(x-1) 
eq 0$ $\Rightarrow x 
eq 1, x 
eq \frac{5}{2}$.$2$. For the logarithm: $x^2-x-2 > 0 \Rightarrow (x-2)(x+1) > 0$. This inequality holds for $x \in (-\infty, -1) \cup (2, \infty)$.Combining these conditions: We need $x \in (-\infty, -1) \cup (2, \infty)$ such that $x 
eq 1$ and $x 
eq \frac{5}{2}$.Since $1$ is not in the interval $(-\infty, -1) \cup (2, \infty)$,we only need to exclude $\frac{5}{2}$.Thus,the domain is $(-\infty, -1) \cup (2, \frac{5}{2}) \cup (\frac{5}{2}, \infty)$.

The domain of the real-valued function $f(x) = \sqrt[3]{\frac{x-2}{2x^2-7x+5}} + \log(x^2-x-2)$ is

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