Q.Find the intervals in which the function $f$ given by $f(x) = 2x^2 - 3x$ is

Concept: Increasing Function Test — A function $f$ is increasing where $f'(x) > 0$ and decreasing where $f'(x) < 0$.

Step 1: Differentiate.

$f'(x) = 4x - 3$.

Step 2: Find the critical point.

Set $f'(x) = 0 \Rightarrow 4x - 3 = 0 \Rightarrow x = \frac{3}{4}$.

Step 3: Test sign on either side.

• For $x < \frac{3}{4}$, say $x = 0$: $f'(0) = -3 < 0$ → decreasing. …

The key idea is simple: the derivative tells you the slope of the tangent at any point. If the slope is positive, the function is rising as you move right — that's increasing. If the slope is negative, it's falling — that's decreasing. This is the Increasing Function Test, and it's the backbone of all such problems.

For a quadratic like $f(x)=2x^2-3x$, the graph is a parabola opening upward (since the coefficient of $x^2$ is positive). So it will decrease until its vertex, then increase after. The derivative will find that turning point exactly.

Let's work through it.

1. Find the derivative.

   $f(x) = 2x^2 - 3x$

   Differentiate term by term:

   $f'(x) = 4x - 3$

2. Set the derivative to zero to find the critical point.

   $4x - 3 = 0 \implies x = \frac{3}{4}$

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

3. Test the sign of $f'(x)$ on either side of $x = \frac{3}{4}$.

   Pick a number less than $\frac{3}{4}$, say $x = 0$:

   $f'(0) = 4(0) - 3 = -3$, which is negative. So $f$ is decreasing on $(-\infty, \frac{3}{4})$.

   Pick a number greater than $\frac{3}{4}$, say $x = 1$: …

Method: Finding Intervals of Increase and Decrease (Sign-Chart Method)

When a question asks to find where a function increases or decreases — rather than prove it increases everywhere — the derivative typically changes sign, so you need to locate the sign-change points and test each resulting interval.

Steps

Step 1: Differentiate the function

Compute $f'(x)$.

Step 2: Find the critical points by solving $f'(x) = 0$

These are the only points where the sign of $f'(x)$ can switch (for a differentiable function). They split the real line into open sub-intervals.

Step 3: Test the sign of $f'(x)$ in each sub-interval

Pick any convenient test value inside each sub-interval (not the critical point itself) and substitute it into $f'(x)$ to see whether the result is positive or negative.

Step 4 (Applying to this problem): Read off increasing/decreasing intervals …

Common Mistakes

Mistake 1: Including the critical point $x=\frac{3}{4}$ in the increasing or decreasing interval

Why it's wrong: At $x=\frac{3}{4}$, the derivative $f'(x)=4x-3$ is exactly zero — the function is neither increasing nor decreasing at that single instant, so it cannot belong to either open interval. Correct approach: always write the intervals as open, $\left(-\infty, \frac{3}{4}\right)$ and $\left(\frac{3}{4}, \infty\right)$, excluding the critical point itself.

Mistake 2: Picking a test point too close to the critical point, risking a sign slip …