Consider the function f(x) = (x - 2) log_e x. | vedclass

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(A) Let $g(x) = x \log_e x + x - 2$. We are looking for roots of the equation $g(x) = 0$. Evaluate $g(x)$ at the endpoints of the interval $(1, 2)$: $

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has at least one root in $(1, 2)$

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(A) Let $g(x) = x \log_e x + x - 2$. We are looking for roots of the equation $g(x) = 0$. Evaluate $g(x)$ at the endpoints of the interval $(1, 2)$: $g(1) = 1 \cdot \log_e(1) + 1 - 2 = 0 + 1 - 2 = -1$. $g(2) = 2 \cdot \log_e(2) + 2 - 2 = 2 \log_e(2) \approx 2(0.693) = 1.386$. Since $g(1) = -1  0$, and $g(x)$ is a continuous function on the interval $[1, 2]$, by the Intermediate Value Theorem, there must exist at least one root $c \in (1, 2)$ such that $g(c) = 0$.

Consider the function $f(x) = (x - 2) \log_e x$. Then the equation $x \log_e x = 2 - x$

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