Lagrange's mean value theorem — geometric interpretation and numerical

Question

State Lagrange’s Mean Value Theorem (LMVT). Verify it for $f(x) = x^3 - 3x + 2$ on the interval $[0, 2]$ . Find the value of $c$ .

(JEE Main 2023, similar pattern)

Solution — Step by Step

Lagrange’s MVT: If $f(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$ , then there exists at least one $c \in (a, b)$ such that:

f'(c) = \frac{f(b) - f(a)}{b - a}

Geometrically: there is at least one point where the tangent is parallel to the secant line joining $(a, f(a))$ and $(b, f(b))$ .

$f(x) = x^3 - 3x + 2$ is a polynomial — it is continuous and differentiable everywhere. So both conditions of LMVT are satisfied on $[0, 2]$ .

f(0) = 0 - 0 + 2 = 2

f(2) = 8 - 6 + 2 = 4

\frac{f(2) - f(0)}{2 - 0} = \frac{4 - 2}{2} = 1

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

Set $f'(c) = 1$ :

3c^2 - 3 = 1

3c^2 = 4

c^2 = \frac{4}{3}

c = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.155

Since $c \approx 1.155 \in (0, 2)$ , LMVT is verified. The value is $c = \frac{2\sqrt{3}}{3}$ .

(We discard $c = -\frac{2\sqrt{3}}{3}$ since it does not lie in $(0, 2)$ .)

Why This Works

Geometrically, the secant line from $(0, 2)$ to $(2, 4)$ has slope 1. The MVT guarantees that somewhere between 0 and 2, the curve’s tangent also has slope 1. At $c = 2\sqrt{3}/3$ , the curve is momentarily climbing at exactly the same rate as the average rate over the whole interval.

Think of it like driving: if you covered 100 km in 2 hours, your average speed was 50 km/h. MVT says at some instant during the trip, your speedometer showed exactly 50 km/h. You must have passed through the average speed at least once.

Alternative Method

You can also verify by graphing: plot $f(x) = x^3 - 3x + 2$ and draw the secant line $y = x + 2$ (slope 1, passing through $(0,2)$ ). Then check visually where the tangent is parallel to this secant — it happens near $x \approx 1.15$ .

JEE Main often asks: “Find the value of $c$ in LMVT for a given function.” It is a straightforward 3-step process: (1) compute the average slope, (2) differentiate, (3) solve $f'(c) = \text{average slope}$ . Just make sure $c$ lies in the open interval $(a, b)$ .

Common Mistake

Students sometimes include the endpoints when checking if $c \in (a, b)$ . The MVT guarantees $c$ in the open interval $(a, b)$ , not the closed interval. If the solution gives $c = a$ or $c = b$ , it does not satisfy the theorem — look for other solutions. Also, there can be multiple valid values of $c$ ; the theorem says “at least one,” not “exactly one.”

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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