Question

75. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.g(x) = x(ln x)²

Accepted Answer

First, identify the domain of the function $$g(x) = x(\ln x)^2$. Since $$\ln x$$ is defined only for x$$ \u003e 0$$, the domain is x$$ \u003e 0$$. Next, find the first derivative g$$'(x)$$ using the product rule. Let u = x$ and v$$ = (\ln $$x)^2. $$Then, $$g'(x) = u'v + uv'. $$Calculate $$u' = 1$$ and $$v' = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}$$, so $$g'(x) = (1)(\ln x)^2$ + x \cdot \frac{2 \ln x}{x} = (\ln x)^2$ + 2 \ln x. $$Set the first derivative equal to zero to find critical points: $$(\ln x)^2$ + 2 \ln x = 0. $$Factor this as $$\ln x (\ln x + 2) = 0$$, which gives $$\ln x = 0 or$$ $$\ln x = -2. $$Solve for $$x to $$find critical points: $$x = e^0$ = 1$$ and $$x = e$$^{-2}$$. Use the second derivative test to classify each critical point. Find g$$''(x)$$ by differentiating g$$'(x) = (\ln $$x)^2 + 2$$ \ln x$$. Differentiate term-by-term: $$\frac{d}{dx} (\ln $$x)^2 = 2$$ \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}$$ and $$\frac{d}{dx} (2 \ln x) = \frac{2}{x}$$. So, g$$''(x) = \frac{2 \ln x}{x} + \frac{2}{x} = \frac{2(\ln x + 1)}{x}$$. Evaluate g$$''(x)$$ at each critical point to determine concavity: if g$$''(x) \u003e 0$$, the point is a local minimum; if g$$''(x) \u003c 0$$, it is a local maximum. Also, consider the behavior of g(x$$)$$ as x$$ 	o 0^+$$ and as x$$ 	o \infty$$ to identify any absolute extrema on the domain.

75. b. Identify the function’s local and absolute extreme values, if any, saying where they occur.
g(x) = x(ln x)²

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Key Concepts

Critical Points and Derivatives

Local and Absolute Extrema

Domain Considerations for Logarithmic Functions