Q.Show that the function $f$ given by $f(x) = x^3 - 3x^2 + 4x,\ x \in \mathbb{R}$ is increasing on $\mathbb{R}$.

Concept: Derivative Sign Analysis — a function is increasing on $\mathbb{R}$ if its derivative is non-negative for all $x$ and zero only at isolated points.

Step 1: Differentiate $f(x) = x^3 - 3x^2 + 4x$:

Step 2: Check the discriminant of $f'(x)$:

Since the coefficient of $x^2$ is positive ($3 > 0$), $f'(x) > 0$ for all real $x$.

Step 3: Because $f'(x) > 0$ everywhere, $f$ is strictly increasing on $\mathbb{R}$.

To show a function is increasing on the whole real line, we need to prove that its derivative is never negative — in fact, strictly positive everywhere. The derivative tells us the slope of the tangent at each point; if that slope is always positive, the function never goes downhill.

Let’s find $f'(x)$.

1. Differentiate term by term.

   $f(x) = x^3 - 3x^2 + 4x$

   Using the power rule:

   $f'(x) = 3x^2 - 6x + 4$

2. Check the sign of this quadratic.

   A quadratic $ax^2 + bx + c$ is always positive for all real $x$ if two conditions hold:

   • $a > 0$ (opens upward)
   • Discriminant $D = b^2 - 4ac < 0$ (no real roots, so it never touches zero)

   Here $a = 3$, $b = -6$, $c = 4$.

   Compute the discriminant:

   $D = (-6)^2 - 4(3)(4) = 36 - 48 = -12$

   Since $D < 0$ and $a = 3 > 0$, the quadratic $3x^2 - 6x + 4$ is positive for every real $x$.

3. Conclude from derivative sign. Since $f'(x) > 0$ for all $x \in \mathbb{R}$, the function $f$ is strictly increasing on $\mathbb{R}$.

Method: Proving a Polynomial's Derivative Never Changes Sign Using the Discriminant

Some "show this cubic (or higher-degree polynomial) is increasing/decreasing on all of $\mathbb{R}$" questions produce a derivative that is itself a quadratic with no real roots — this method proves that quadratic never crosses zero, without needing a sign chart at all.

Steps

Step 1: Differentiate to get $f'(x)$.

If $f(x)$ is a cubic, $f'(x)$ will be a quadratic $ax^2+bx+c$.

Step 2: Compute the discriminant of $f'(x)$.

Step 3: Interpret the discriminant together with the leading coefficient.

If $D < 0$, the quadratic $f'(x)$ has no real roots, so it never touches zero — it keeps one constant sign for every real $x$. The sign itself is decided by the leading coefficient $a$: if $a>0$ the quadratic (and hence $f'(x)$) is positive for all $x$; if $a<0$ it is negative for all $x$.

Step 4: Complete the square as an extra check (optional but reassuring).

Writing $f'(x)$ as $a(x-h)^2+k$ with $k>0$ (when $a>0$) makes the "always positive" claim visually obvious and is a good way to double-check the discriminant argument.

Common Mistakes

Mistake 1: Checking the derivative's sign at only a few sample points instead of proving it for every real $x$.

Why it's wrong: plugging in a handful of values and seeing $f'(x)>0$ each time does not rule out the derivative turning negative somewhere you didn't check — a "show that" question demands a complete argument, not spot-checks. Correct approach: use the discriminant of the quadratic $f'(x)$ to prove algebraically that it never touches zero for any real $x$.

Mistake 2: Concluding "always positive" from a negative discriminant alone, without checking the leading coefficient.

Why it's wrong: a negative discriminant only guarantees the quadratic never crosses zero — it does not tell you which constant sign it holds. A quadratic with $D<0$ and a negative leading coefficient is negative for every $x$, not positive. Correct approach: state both conditions together — $D<0$ and $a>0$ — before concluding $f'(x)>0$ everywhere.