Q.Prove that the function given by $f(x) = x^3 - 3x^2 + 3x - 100$ is increasing in $\mathbf{R}$.

Concept: Mean Value Theorem — a function with a positive derivative everywhere is strictly increasing.

Step 1: Differentiate $f(x)$:

Step 2: For all real $x$, $(x-1)^2 \ge 0$, so $f'(x) = 3(x-1)^2 \ge 0$.

Equality holds only at $x = 1$, but the derivative is never negative.

Step 3: By the Mean Value Theorem, if $a < b$, there exists $c \in (a,b)$ such that

$f(b)-f(a) = f'(c)(b-a) \ge 0,$ …

To test monotonicity we examine the sign of the derivative: if $f'(x) \ge 0$ throughout an interval (with equality only at isolated points), then $f$ is increasing there.

1. Differentiate.

2. Determine the sign.

Since $(x-1)^2 \ge 0$ for every real $x$,

with equality only at the single point $x = 1$.

3. Conclude. …

Method: Proving a Function Is Increasing Everywhere Using a Perfect-Square Derivative

Use this method whenever a cubic (or other polynomial) is claimed to be increasing on the whole real line — the standard trick is to show the derivative factors into a perfect square (or sum of squares), which is automatically non-negative.

Steps

Step 1: Differentiate the function.

Compute $f'(x)$, which will typically be a quadratic for a cubic $f$.

Step 2: Try to factor the quadratic derivative into a perfect square.

If $f'(x)$ can be written as $k(x-c)^2$ for a positive constant $k$, this is the key structural fact the whole proof relies on — a zero discriminant on the quadratic signals that a perfect square is available.

Step 3: Argue that $f'(x)\ge 0$ for every real $x$, with equality only at the single isolated point $x=c$.

$(x-c)^2\ge 0$ always, so $k(x-c)^2\ge 0$ for $k>0$, equalling zero only at the single point $x=c$ and nowhere else. …

Common Mistakes

Mistake 1: Concluding the function is not strictly increasing because $f'(x)=0$ at one point ($x=1$).

Why it's wrong: a derivative touching zero at an isolated point (not throughout an interval) does not break strict monotonicity — the function still climbs continuously through that point, it just has a momentary horizontal tangent. Correct approach: distinguish "zero at one point" from "zero throughout an interval"; only the latter would prevent strict increase.

Mistake 2: Testing the sign of $f'(x)=3x^2-6x+3$ with a handful of sample points instead of factoring it. …