Q.Find the intervals in which the function $f$ given by $f(x) = x^2 - 4x + 6$ is

Concept: Monotonic Function Analysis

A function increases where $f'(x) > 0$ and decreases where $f'(x) < 0$.

Step 1: Find the derivative

$f(x) = x^2 - 4x + 6$

$f'(x) = 2x - 4$

Step 2: Find the critical point

Set $f'(x) = 0$:

$2x - 4 = 0 \implies x = 2$

Step 3: Test intervals

• For $x < 2$, say $x = 0$: $f'(0) = -4 < 0$ → decreasing …

The core idea here is monotonic function analysis using the first derivative. For a differentiable function, the sign of $f'(x)$ tells us everything about where the function is increasing or decreasing.

Why does this work? The derivative $f'(x)$ measures the instantaneous rate of change. If $f'(x) > 0$ over an interval, the function's value rises as $x$ increases — that's an increasing function. If $f'(x) < 0$, the value falls — that's a decreasing function. The points where $f'(x) = 0$ are potential turning points (where the monotonicity can change).

For a quadratic like $f(x) = x^2 - 4x + 6$, the graph is a parabola opening upward (since the coefficient of $x^2$ is positive). So it must have a single minimum at its vertex, decreasing before it and increasing after. The derivative approach will confirm this and give us the exact intervals.

Let's work through it step by step.

1. Find the derivative.

   $f(x) = x^2 - 4x + 6$

   Differentiating term by term:

   $f'(x) = 2x - 4$

2. Find the critical point.

   Set $f'(x) = 0$:

   $2x - 4 = 0 \implies x = 2$

   This is the only point where the derivative changes sign — the vertex of the parabola.

3. Test the sign of $f'(x)$ on either side of $x=2$.

   Pick a test point to the left, say $x = 0$:

   $f'(0) = 2(0) - 4 = -4 < 0$

   So $f'(x) < 0$ for all $x < 2$ (since the derivative is linear and crosses zero only once).

   Pick a test point to the right, say $x = 3$:

   $f'(3) = 2(3) - 4 = 2 > 0$ …

Method: The First-Derivative Sign-Chart Test for Intervals of Increase/Decrease

This is the general method for "find the intervals in which $f$ is increasing/decreasing" — unlike "show $f$ is increasing everywhere," this kind of question expects the domain split into pieces.

Steps

Step 1: Differentiate.

Find $f'(x)$.

Step 2: Solve $f'(x)=0$ to locate the critical points.

These are the only places the sign of $f'(x)$ can change (for a polynomial, the sign stays fixed between consecutive real roots).

Step 3: Split the domain into intervals at the critical points, and test the sign of $f'(x)$ in each.

Pick any convenient value inside each interval and substitute into $f'(x)$ — only the sign matters, not the exact value. …

Common Mistakes

Mistake 1: Making a sign error when substituting a test point into the derivative.

Why it's wrong: a slip like computing $f'(0) = 2(0)-4 = 4$ instead of $-4$ flips the entire conclusion about which side of the critical point is increasing and which is decreasing. Correct approach: substitute carefully and double-check the arithmetic sign of the result, not just its magnitude.

Mistake 2: Excluding the critical point from both intervals, writing $(-\infty,2)$ and $(2,\infty)$ instead of the closed-interval form. …