Q.Find the value of $c$ from Lagrange's Mean Value Theorem for the function $f(x) = x^2 - 1$ on $[2, 3]$.

The Mean Value Theorem

Imagine you drive Delhi to Agra — 200 km — in exactly 4 hours, so your average speed is 50 km/h. Was your speed exactly 50 km/h at some instant? If your motion was smooth, the Mean Value Theorem says yes. That is its soul: it links the average rate of change of a function over an interval to its instantaneous rate at some point inside.

The Intuition

Think of $f(x)$ as a smooth path from $x = a$ to $x = b$. The average rate of change is the slope of the chord joining the endpoints:

If the path has no sharp corners or breaks, then at some interior point the tangent's slope must exactly equal this chord slope — geometrically, the tangent there is parallel to the chord. (If you always went slower than average you'd never arrive; always faster and you'd overshoot — so you must hit the average at least once.)

The Precise Statement

Continuity means no breaks; differentiability means a well-defined tangent at every interior point — no corners, no vertical tangents.

A Simple Example

Take $f(x) = x^2$ on $[1, 3]$. The average slope is $\dfrac{9 - 1}{3 - 1} = 4$, and $f'(x) = 2x$. Setting $2c = 4$ gives $c = 2 \in (1, 3)$, and indeed $f'(2) = 4$.

Why It Matters …