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The domain of a function is the set of all possible input values (or \(x\)-values) that the function can accept. It's crucial to determine the domain, especially in functions involving logarithms, radicals, or fractions, as these have specific constraints. For the function \(f(x) = \ln(x^2 + x)\), we need to find when the argument of the logarithm, \(x^2 + x\), is greater than zero because logarithms are only defined for positive numbers.

To solve the inequality \(x^2 + x > 0\), you can factor it as \(x(x + 1) > 0\). This inequality implies two critical conditions:

• Either both factors are positive: \(x > 0\) and \(x + 1 > 0\), meaning \(x > 0\).
• Or both factors are negative: \(x < 0\) and \(x + 1 < 0\), meaning \(x < -1\).

Consequently, the domain of the function \(f(x)\) is \((-\infty, -1) \cup (0, \infty)\). For plotting purposes within the given window of \([-10, 10]\), the domain is \([-10, -1) \cup (0, 10]\). Avoid the points where \(x\) would lead to zero in the argument of the logarithm as these make the function undefined.

Critical points are values of \(x\) where the derivative of a function equals zero or is undefined. These points help identify potential maxima, minima, or points of inflection on the function's graph.

For our function \(f(x) = \ln(x^2 + x)\), the derivative is \(f'(x) = \frac{2x + 1}{x^2 + x}\). To find critical points, set \(f'(x) = 0\). This simplifies to \(2x + 1 = 0\). Solving this equation yields \(x = -\frac{1}{2}\). This point is within the interval \([-10, -1)\), making it a valid critical point for our analysis.

Additionally, the derivative may be undefined when \(x = 0\) or \(x =-1\), but these values are, in fact, outside the defined domain of the function. Hence, our focus remains on \(x = -\frac{1}{2}\) to evaluate the changes in function behavior
.

Increasing and decreasing intervals tell us about the behavior of the function over specific sections. If the derivative \(f'(x) > 0\) in an interval, the function is increasing there; whereas if \(f'(x) < 0\), it is decreasing.

Around our critical point \(x = -\frac{1}{2}\), test the sign of the derivative in adjacent intervals to determine where the function increases or decreases:

• For \(x < -\frac{1}{2}\) (a number like \(-1\) was tested), \(f'(x) > 0\). Therefore, the function is increasing in \([-10, -\frac{1}{2})\).
• Exactly at \(x = -\frac{1}{2}\), the behavior changes, as \(f'(x) < 0\) for \(-\frac{1}{2} < x < -1\). Thus, this interval \([-\frac{1}{2}, -1)\) shows a decreasing behavior.
• For intervals in \((0, 10]\), the derivative \(f'(x) > 0\), as tested with \(x = 1\), confirms the function is increasing.

Identifying these intervals not only helps in sketching the graph but also gives insight into the function's overall behavior: increasing on \([-10, -\frac{1}{2})\) and \((0, 10]\), while decreasing on \([-\frac{1}{2}, -1)\). Understanding these patterns ensures a clear grasp of how the function behaves across its domain.