Q.If $M$ and $m$ denote the local maximum and local minimum values of the function $f(x) = x + \dfrac{1}{x}$ $(x \ne 0)$ respectively, find the value of $(M - m)$.

Local Maximum Condition

Imagine hiking in a mountain range. You reach a point where, for a few steps in any direction, the ground drops away. You may not be the highest peak in the whole range, but right here every step goes downhill. That's a local maximum: a point higher than all nearby points.

The Intuition First

On a smooth, wavy curve, a local maximum is a "hilltop" — a point where the function peaks, then falls on both sides. Zoom in and the values just to the left and right are both lower.

"Local" means "in a small neighbourhood." The function might have higher values elsewhere (a global maximum), but that doesn't matter — a local maximum is king of its own tiny kingdom.

The Precise Mathematical Statement

Let $f$ be a real-valued function on a domain $D \subseteq \mathbb{R}$. A point $c \in D$ is a local maximum if there exists some $\delta > 0$ such that for every $x$ in the domain within distance $\delta$ of $c$:

That is, on some open interval $(c - \delta, c + \delta)$, $f(c)$ is the largest value.

The First Derivative Test

If $f$ is differentiable at $c$ and $c$ is a local maximum, then:

This is the critical point condition — the tangent is horizontal. But $f'(c) = 0$ is necessary, not sufficient: a horizontal tangent could also be a local minimum or a saddle point (like $f(x) = x^3$ at $x = 0$).

The Second Derivative Test

If $f'(c) = 0$ and $f''(c) < 0$, then $c$ is a local maximum: a negative second derivative means $f$ is concave down at $c$ — curving down like an upside-down bowl. If $f''(c) > 0$, it's a local minimum. If $f''(c) = 0$, the test is inconclusive.

A Concrete Example

Consider $f(x) = -x^2 + 4x - 3$. …