Q.Find the interval in which the function $f(x)=10-6x-2x^2$ is strictly increasing or strictly decreasing.

Derivative Sign Analysis: What the Slope Tells You

Imagine walking along a hilly road — sometimes uphill, sometimes downhill, occasionally flat. The derivative at any point is simply the slope of the road under your feet at that instant.

Derivative sign analysis figures out where a function is increasing, where it is decreasing, and where it has flat spots (critical points) — all from the sign of its derivative.

The Intuition First

If $f'(x)$ is positive, the function is increasing — the graph rises as you move right. If $f'(x)$ is negative, it is decreasing. If $f'(x) = 0$, there is a horizontal tangent — a potential peak, valley, or flat inflection.

The key: a single point tells you little; you look at intervals. If $f'(x) > 0$ for all $x$ in $(a, b)$, the function is strictly increasing on that whole interval. Same logic for negative.

The Precise Statement

Let $f$ be differentiable on an open interval $I$. Then:

• If $f'(x) > 0$ for all $x$ in $I$, then $f$ is strictly increasing on $I$.
• If $f'(x) < 0$ for all $x$ in $I$, then $f$ is strictly decreasing on $I$.
• If $f'(x) = 0$ for all $x$ in $I$, then $f$ is constant on $I$.

Points where $f'(x) = 0$ (or where $f'$ does not exist) are critical points — the candidates for local maxima and minima.

How to Perform It (Step-by-Step)

1. Find the derivative $f'(x)$.
2. Find critical points: solve $f'(x) = 0$ and check where $f'(x)$ is undefined (but $f$ is defined).
3. Plot these on a number line — they split the domain into intervals.
4. Pick a test point inside each interval and evaluate $f'$; only the sign matters.
5. Record the sign in each interval and interpret: + means increasing, – means decreasing.

A Concrete Example

Take $f(x) = x^3 - 3x$.

Step 1: $f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$.

Step 2: Critical points: $x = -1$ and $x = 1$.

Step 3: Intervals: $(-\infty, -1)$, $(-1, 1)$, $(1, \infty)$.

Step 4: Test points:

• $x = -2$: $f'(-2) = 3(4-1) = 9 > 0$.
• $x = 0$: $f'(0) = -3 < 0$.
• $x = 2$: $f'(2) = 9 > 0$.

Step 5: So $f$ increases on $(-\infty, -1)$, decreases on $(-1, 1)$, increases on $(1, \infty)$. Thus $x = -1$ is a local maximum (sign changes + to –), and $x = 1$ is a local minimum (– to +). …