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(B) Given the function $f(x) = x(\ln x - 2)$ on the interval $[1, e^2]$.To find the critical points,we find the derivative $f'(x)$:$f'(x) = x \cdot \f

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$e$

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(B) Given the function $f(x) = x(\ln x - 2)$ on the interval $[1, e^2]$.To find the critical points,we find the derivative $f'(x)$:$f'(x) = x \cdot \frac{d}{dx}(\ln x - 2) + (\ln x - 2) \cdot \frac{d}{dx}(x)$$f'(x) = x \cdot \frac{1}{x} + \ln x - 2 = 1 + \ln x - 2 = \ln x - 1$.Setting $f'(x) = 0$ gives $\ln x - 1 = 0$,so $\ln x = 1$,which implies $x = e$.Now,we evaluate the function at the critical point and the endpoints of the interval:$f(1) = 1(\ln 1 - 2) = 1(0 - 2) = -2$.$f(e) = e(\ln e - 2) = e(1 - 2) = -e$.$f(e^2) = e^2(\ln e^2 - 2) = e^2(2 - 2) = 0$.Comparing the values $\{-2, -e, 0\}$,since $e \approx 2.718$,the least value is $-e$ and the greatest value is $0$.The difference between the greatest and the least values is $0 - (-e) = e$.

The difference between the greatest and the least values of the function $f(x) = x(\ln x - 2)$ on the interval $[1, e^2]$ is:

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