The function \(f(x) = x^2-x+1\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → Increasing on \(\left(\displaystyle\frac{1}{2},\infty\right)\) and decreasing on \(\left(-\infty,\displaystyle\frac{1}{2}\right)\)
The sign of the derivative decides where f increases or decreases:
\(f'(x) = 2x - 1\)
\(\displaystyle f'(x) = 0 \text{ at } x = \frac{1}{2}\)
For \(x \lt \frac{1}{2}\), \(f'(x) \lt 0\), so f is decreasing on \(\left(-\infty, \frac{1}{2}\right)\).
For \(x \gt \frac{1}{2}\), \(f'(x) \gt 0\), so f is increasing on \(\left(\frac{1}{2}, \infty\right)\).
Why the other options are incorrect
Option (1) Increasing on \(\left(\frac{-1}{2},1\right)\) and decreasing on \(\left(0,\frac{1}{2}\right)\) — Incorrect. f is decreasing on \(\left(-\frac{1}{2}, \frac{1}{2}\right)\), which is part of \(\left(-\frac{1}{2}, 1\right)\); for example, f(0) = 1 is less than f(−1/2) = 7/4.
Option (3) Increasing on \(\left(-\infty,\frac{1}{2}\right]\) and decreasing on \(\left[\frac{1}{2},\infty\right)\) — Incorrect. This is the behaviour reversed; it would be true for \(-f\), whose derivative is \(1 - 2x\).
Option (4) Increasing on \((-\infty, 1)\) and decreasing on \((1,\infty)\) — Incorrect. The turning point is at \(x = \frac{1}{2}\), not 1, and f is increasing, not decreasing, for x > 1.