Discussion
If \(g(x) = x^{2} - x, x \in \mathbb{R}\), then \(g(x)\) is increasing in
Answer: E
g is increasing in \(\left[\frac{1}{2}, \infty\right)\).
\(g'(x) = 2x - 1\), which is non-negative for \(x \ge \frac{1}{2}\), so g increases on \(\left[\frac{1}{2}, \infty\right)\). It decreases on \(\left(-\infty, \frac{1}{2}\right]\), with its minimum at \(x = \frac{1}{2}\).
\((-\infty, \infty)\) and (−5, 5) include part of the decreasing stretch, and \((-\infty, 0)\) lies wholly in it. '\((0, -\infty)\)' is not a valid interval.
A quadratic \(ax^{2} + bx + c\) with \(a \gt 0\) always decreases up to its vertex at \(x = -\frac{b}{2a}\) and increases after it; here the vertex is at \(x = \frac{1}{2}\).
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