The equation x log x = 3 - x: | vedclass

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(B) Let $f(x) = x \log x + x - 3$. $f'(x) = x \cdot \frac{1}{x} + \log x + 1 = 1 + \log x + 1 = \log x + 2$. For $x \in (1, 3)$, $\log x > 0$, so $f'(

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has exactly one root in $(1, 3)$

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(B) Let $f(x) = x \log x + x - 3$. $f'(x) = x \cdot \frac{1}{x} + \log x + 1 = 1 + \log x + 1 = \log x + 2$. For $x \in (1, 3)$, $\log x > 0$, so $f'(x) = \log x + 2 > 2 > 0$. Thus, $f(x)$ is strictly increasing on $(1, 3)$. $f(1) = 1 \cdot \log(1) + 1 - 3 = 0 + 1 - 3 = -2$. $f(3) = 3 \log 3 + 3 - 3 = 3 \log 3 \approx 3(1.098) = 3.294 > 0$. Since $f(1)  0$ and $f(x)$ is continuous and strictly increasing, by the Intermediate Value Theorem, there exists exactly one root in $(1, 3)$.

The equation $x \log x = 3 - x$:

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