To study the sign of the function $g(x) = (x-1)^2 (\ln{x} - 1)$, you can analyze it without differentiating by focusing on the behavior of each factor. The function is composed of two parts: $(x - 1)^2$ and $(\ln{x} - 1)$. To determine the sign of the product, it is enough to evaluate the sign of each part and combine them.

Step-by-Step Analysis:

1. Behavior of $(x - 1)^2$:

   • $(x - 1)^2$ is a square, so it is always non-negative: $(x - 1)^2 \geq 0 \quad \text{for all} \ x.$
   • $(x - 1)^2 = 0$ only when $x = 1$. Therefore, this factor is zero at $x = 1$ and positive everywhere else.

2. Behavior of $\ln{x} - 1$:

   • The logarithmic function $\ln{x}$ is defined for $x > 0$.
   • For $x = e$, $\ln{e} = 1$, so $\ln{e} - 1 = 0$. Therefore, $\ln{x} - 1 = 0$ at $x = e$.
   • For $0 < x < e$, $\ln{x} < 1$, so $\ln{x} - 1 < 0$.
   • For $x > e$, $\ln{x} > 1$, so $\ln{x} - 1 > 0$.

Sign of $g(x) = (x - 1)^2 (\ln{x} - 1)$:

• For $x = 1$, $g(1) = (1 - 1)^2 (\ln{1} - 1) = 0$, so $g(1) = 0$.
• For $0 < x < 1$, $(x - 1)^2 > 0$ and $\ln{x} - 1 < 0$, so $g(x) < 0$.
• For $1 < x < e$, $(x - 1)^2 > 0$ and $\ln{x} - 1 < 0$, so $g(x) < 0$.
• For $x = e$, $g(e) = (e - 1)^2 (\ln{e} - 1) = (e - 1)^2 \cdot 0 = 0$, so $g(e) = 0$.
• For $x > e$, $(x - 1)^2 > 0$ and $\ln{x} - 1 > 0$, so $g(x) > 0$.

Why Differentiation is Not Needed:

To study the sign, you don't need to differentiate because the function's sign depends only on the sign of each factor. You can analyze each factor separately and combine the results, as shown above.

Would you like more details on any part of this explanation?

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Related Questions:

1. What are the key points for studying the sign of a product of functions?
2. How does the behavior of $\ln{x}$ change for different values of $x$?
3. What role does the factor $(x - 1)^2$ play in determining the sign of $g(x)$?
4. What is the significance of the points $x = 1$ and $x = e$ in the function?
5. How does the square function affect the sign of a product when combined with a changing function?

Tip:

When analyzing a product of functions, you can often determine the sign by studying the sign of each factor separately, avoiding unnecessary differentiation.