Apply Lagrange MVT: f'(c) = (f(b)-f(a))/(b-a).

Mean Value Theorem — Find c for f(x) = x³ on [0,2]

Question

Find the value of $c$ guaranteed by Lagrange’s Mean Value Theorem for $f(x) = x^3$ on the interval $[0, 2]$ .

Solution — Step by Step

MVT requires $f$ to be continuous on $[0, 2]$ and differentiable on $(0, 2)$ . Since $f(x) = x^3$ is a polynomial, both conditions are satisfied — no issues here.

With $a = 0$ and $b = 2$ :

f(0) = 0^3 = 0, \quad f(2) = 2^3 = 8

The average rate of change over $[0, 2]$ is:

\frac{f(b) - f(a)}{b - a} = \frac{8 - 0}{2 - 0} = \frac{8}{2} = 4

MVT says there exists at least one $c \in (0, 2)$ such that $f'(c)$ equals the average rate of change. We need $f'(x)$ first:

f'(x) = 3x^2 \implies f'(c) = 3c^2

Now set this equal to 4:

3c^2 = 4

c^2 = \frac{4}{3} \implies c = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}

We need $c \in (0, 2)$ , so we discard the negative root.

\boxed{c = \frac{2}{\sqrt{3}} = \frac{2\sqrt{3}}{3} \approx 1.15}

Check: $\frac{2\sqrt{3}}{3} \approx 1.15$ , which lies in $(0, 2)$ . ✓

Why This Works

MVT is essentially saying this: if you drove from Delhi to Agra at an average speed of 80 km/h, then at some moment during the journey your instantaneous speed was exactly 80 km/h. The average rate of change is the slope of the secant line joining $(a, f(a))$ and $(b, f(b))$ . MVT guarantees a point where the tangent is parallel to that secant.

For $f(x) = x^3$ , the secant from $(0, 0)$ to $(2, 8)$ has slope 4. The parabola $f'(x) = 3x^2$ grows from 0 to 12 on this interval, so it must cross 4 somewhere — that crossing point is our $c$ .

This is why the theorem needs the interval to be closed for continuity but open for differentiability — the endpoints can be sharp corners, but the interior must be smooth.

Alternative Method

Rather than rationalizing, you can leave the answer as $c = \dfrac{2}{\sqrt{3}}$ and verify numerically. Plug back in:

f'(c) = 3 \cdot \left(\frac{2}{\sqrt{3}}\right)^2 = 3 \cdot \frac{4}{3} = 4 \checkmark

This back-substitution check takes 10 seconds and will save marks in CBSE boards — many students lose a step mark for not verifying.

CBSE 2025 Sample Paper carried this exact question for 3 marks. The mark scheme gives 1 mark each for: checking conditions, setting up $f'(c) = \frac{f(b)-f(a)}{b-a}$ , and finding $c$ . Don’t skip step 1 — conditions always carry a mark.

Common Mistake

The most common error: students write $c = \pm \dfrac{2}{\sqrt{3}}$ and report both values as the answer. MVT asks for $c \in (a, b) = (0, 2)$ , so $c = -\dfrac{2}{\sqrt{3}} \approx -1.15$ is outside the interval and must be rejected. In CBSE marking, reporting the negative root without rejection typically costs you the final mark.

A second trap: forgetting to rationalize. CBSE expects $\dfrac{2\sqrt{3}}{3}$ , not $\dfrac{2}{\sqrt{3}}$ . Both are mathematically correct, but rationalized form is the convention for full marks in board exams.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

Try These Next