The set of values of x for which the function | vedclass

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(C) The function $f(x) = \log \left(\frac{x-1}{x+2}\right)$ is defined and continuous where the argument of the logarithm is strictly positive. We req

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$(-\infty, -2) \cup (1, \infty)$

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(C) The function $f(x) = \log \left(\frac{x-1}{x+2}\right)$ is defined and continuous where the argument of the logarithm is strictly positive. We require $\frac{x-1}{x+2} > 0$. To solve this inequality, we find the critical points by setting the numerator and denominator to zero: $x-1 = 0 \Rightarrow x = 1$ and $x+2 = 0 \Rightarrow x = -2$. Using the wavy curve method (sign scheme) on the number line: For $x > 1$, $\frac{x-1}{x+2} > 0$. For $-2 For $x  0$. Thus, the function is continuous for $x \in (-\infty, -2) \cup (1, \infty)$.

The set of values of $x$ for which the function $f(x) = \log \left(\frac{x-1}{x+2}\right)$ is continuous, is

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