Q.Prove that the function $f$ given by $f(x) = x^2 - x + 1$ is neither strictly increasing nor decreasing on $(-1, 1)$.

Concept: Monotonic Function Analysis — A function is strictly increasing if $f'(x) > 0$ and strictly decreasing if $f'(x) < 0$ on an interval. If the derivative changes sign, the function is neither.

Step 1: Compute the derivative:

$f'(x) = 2x - 1$.

Step 2: Find where $f'(x) = 0$:

$2x - 1 = 0 \implies x = \frac{1}{2}$, which lies inside $(-1, 1)$.

Step 3: Test sign of $f'(x)$ on either side of $x = \frac{1}{2}$:

• For $x < \frac{1}{2}$, say $x = 0$: $f'(0) = -1 < 0$ → decreasing.
• For $x > \frac{1}{2}$, say $x = 1$: $f'(1) = 1 > 0$ → increasing. …

The core idea here is monotonic function analysis: a function is strictly increasing on an interval if its derivative is positive everywhere on that interval, and strictly decreasing if the derivative is negative everywhere. If the derivative changes sign inside the interval, the function cannot be monotonic in either direction.

For $f(x) = x^2 - x + 1$, the derivative is $f'(x) = 2x - 1$. This is a linear function — it crosses zero at exactly one point. That single sign change is the reason the function's behaviour flips.

1. Find the critical point. Set $f'(x) = 0$:

This point lies inside $(-1,1)$ because $-1 < \frac12 < 1$.

2. Check the sign of $f'(x)$ on either side of $x = \frac12$.

   • For $x < \frac12$, say $x = 0$: $f'(0) = -1 < 0$. So $f$ is strictly decreasing on $(-1, \frac12)$.
   • For $x > \frac12$, say $x = 1$ (which is in $(-1,1)$): $f'(1) = 1 > 0$. So $f$ is strictly increasing on $(\frac12, 1)$.

3. Interpret the result.

   Since the function decreases on the left part of $(-1,1)$ and increases on the right part, it is not strictly increasing overall (because it goes down first) and not strictly decreasing overall (because it goes up later). The change in monotonicity at $x = \frac12$ is the key. …

Method: Showing a Function Is Neither Increasing nor Decreasing on an Interval

Use this method whenever a question asks you to prove a function is "neither increasing nor decreasing" on a given interval — this happens exactly when a critical point of the derivative lies strictly inside that interval.

Steps

Step 1: Differentiate the function.

Compute $f'(x)$ using the standard differentiation rules.

Step 2: Solve $f'(x) = 0$ to find the critical point(s).

gives the value(s) of $x$ where the slope is momentarily zero — a candidate turning point.

Step 3: Check whether the critical point lies strictly inside the given interval.

If it does, the derivative (being linear here) necessarily has one sign on one side of that point and the opposite sign on the other, so the function cannot keep the same direction across the whole interval. …

Common Mistakes

Mistake 1: Comparing only the two endpoint values $f(-1)$ and $f(1)$.

Why it's wrong: endpoint comparison only tells you the net change over the whole interval, not whether the function moved in one direction throughout — a function can dip down and rise back up to a higher value, which endpoint comparison completely misses. Correct approach: always test the sign of $f'(x)$ across the interval, not just compare $f$ at the two ends.

Mistake 2: Forgetting to check whether the critical point actually lies inside the stated interval.

Why it's wrong: if $x=\tfrac12$ had fallen outside $(-1,1)$, the derivative would keep one sign throughout and the function would in fact be monotonic there — the "neither" conclusion depends entirely on the critical point being interior to the interval. Correct approach: explicitly verify $-1 < \tfrac12 < 1$ before drawing any conclusion. …