75. a. Find the open intervals on which the function is increasing and decreasing.
g(x) = x(ln x)²

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Identify the domain of the function \(g(x) = x(\ln x)^2\). Since \(\ln x\) is defined only for \(x > 0\), the domain is \((0, \infty)\).

Find the first derivative \(g'(x)\) using the product rule. Let \(u = x\) and \(v = (\ln x)^2\). Then, \(g'(x) = u'v + uv'\). Compute \(u' = 1\) and \(v' = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}\).

Substitute these into the derivative: \(g'(x) = 1 \cdot (\ln x)^2 + x \cdot \frac{2 \ln x}{x} = (\ln x)^2 + 2 \ln x\).

Factor the derivative to simplify: \(g'(x) = \ln x \left( \ln x + 2 \right)\).

Determine the critical points by setting \(g'(x) = 0\), which gives \(\ln x = 0\) or \(\ln x = -2\). Solve these to find \(x = 1\) and \(x = e^{-2}\). Use these points to test intervals in the domain \((0, \infty)\) to find where \(g'(x)\) is positive (increasing) or negative (decreasing).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Domain of the Function

Before analyzing increasing or decreasing behavior, determine the domain where the function is defined. For g(x) = x(ln x)², the natural logarithm ln x requires x > 0, so the domain is (0, ∞). Understanding the domain ensures correct interval analysis.

First Derivative and Critical Points

The first derivative g'(x) indicates the slope of the function. Finding g'(x) and solving g'(x) = 0 identifies critical points where the function may change from increasing to decreasing or vice versa. These points help partition the domain into intervals for testing.

Increasing and Decreasing Intervals

A function is increasing where its derivative is positive and decreasing where its derivative is negative. By testing values in intervals determined by critical points, one can classify each interval as increasing or decreasing, providing a complete description of the function's behavior.

75. a. Find the open intervals on which the function is increasing and decreasing.g(x) = x(ln x)²