Mean value theorem — verify for f(x) = x³ - x on [-1, 1]

Question

Verify the Mean Value Theorem for $f(x) = x^3 - x$ on the interval $[-1, 1]$ . Find all values of $c$ guaranteed by the theorem.

(JEE Main 2021)

Solution — Step by Step

The Mean Value Theorem requires:

$f(x)$ is continuous on $[-1, 1]$ — yes, polynomials are continuous everywhere ✓
$f(x)$ is differentiable on $(-1, 1)$ — yes, polynomials are differentiable everywhere ✓

Both conditions satisfied, so MVT applies.

MVT states there exists $c \in (-1, 1)$ such that:

f'(c) = \frac{f(1) - f(-1)}{1 - (-1)}

$f(1) = 1 - 1 = 0$

$f(-1) = -1 - (-1) = -1 + 1 = 0$

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

f'(x) = 3x^2 - 1

Setting $f'(c) = 0$ :

3c^2 - 1 = 0

c^2 = \frac{1}{3}

c = \pm\frac{1}{\sqrt{3}}

$c = \frac{1}{\sqrt{3}} \approx 0.577$ and $c = -\frac{1}{\sqrt{3}} \approx -0.577$ .

Both values lie in $(-1, 1)$ ✓.

So MVT is verified with $c = \pm\frac{1}{\sqrt{3}}$ .

Why This Works

Geometrically, MVT says: between any two points on a smooth curve, there’s at least one point where the tangent is parallel to the secant (the line joining the two endpoints). Here, the secant from $(-1, 0)$ to $(1, 0)$ has slope $0$ (horizontal). So MVT guarantees a point where $f'(c) = 0$ — a horizontal tangent.

The function $f(x) = x^3 - x$ goes up, comes down, goes up again between $-1$ and $1$ . The two turning points (where the tangent is horizontal) are exactly at $c = \pm 1/\sqrt{3}$ . We get two values of $c$ — MVT only guarantees at least one, but there can be more.

Alternative Method — Rolle’s Theorem

Since $f(-1) = f(1) = 0$ , this is actually a special case of Rolle’s Theorem (MVT with equal endpoint values). Rolle’s Theorem directly says: there exists $c \in (-1, 1)$ with $f'(c) = 0$ .

Rolle’s Theorem is MVT with the extra condition $f(a) = f(b)$ . Whenever $f(a) = f(b)$ , use Rolle’s Theorem directly — it simplifies the right side to zero. JEE questions often test whether you can recognise when Rolle’s applies as a special case of MVT.

Common Mistake

Students sometimes check $c = \pm 1/\sqrt{3}$ but then write only one of them as the answer. MVT guarantees “at least one $c$ ” — but the question asks to “find all values of $c$ .” When the equation gives multiple solutions in the interval, report all of them. Missing a solution means incomplete answer.

Question

Solution — Step by Step

Why This Works

Alternative Method — Rolle’s Theorem

Common Mistake

Try These Next