Check conditions: continuous, differentiable, f(a)=f(b). Find c.

Rolle's Theorem — Verify for f(x) = x² - 4x + 3 on [1,3]

Question

Verify Rolle’s Theorem for $f(x) = x^2 - 4x + 3$ on the interval $[1, 3]$ , and find the value of $c$ guaranteed by the theorem.

Solution — Step by Step

$f(x) = x^2 - 4x + 3$ is a polynomial. Polynomials are continuous everywhere on $\mathbb{R}$ , so $f$ is continuous on $[1, 3]$ . First condition: satisfied.

Polynomials are differentiable everywhere. So $f$ is differentiable on the open interval $(1, 3)$ . Second condition: satisfied.

Calculate the endpoint values:

f(1) = (1)^2 - 4(1) + 3 = 1 - 4 + 3 = 0

f(3) = (3)^2 - 4(3) + 3 = 9 - 12 + 3 = 0

Since $f(1) = f(3) = 0$ , the third condition is satisfied. All three conditions hold, so Rolle’s Theorem applies.

The theorem guarantees at least one $c \in (1, 3)$ where $f'(c) = 0$ .

Differentiate: $f'(x) = 2x - 4$

Set $f'(c) = 0$ :

2c - 4 = 0 \implies c = 2

Since $2 \in (1, 3)$ , this is valid.

The value $c = 2$ satisfies Rolle’s Theorem.

Why This Works

Rolle’s Theorem is essentially saying: if a continuous, smooth curve starts and ends at the same height, somewhere in between it must have a horizontal tangent — it had to turn around.

Geometrically, $f(x) = x^2 - 4x + 3 = (x-1)(x-3)$ is a upward parabola with roots at $x = 1$ and $x = 3$ . The vertex of any parabola has a horizontal tangent, and here the vertex falls exactly at $x = 2$ , which is the midpoint of $[1, 3]$ .

The derivative $f'(x) = 2x - 4$ measures the slope. Setting it to zero locates the flat point. Rolle’s Theorem is the theoretical guarantee that this point must exist — we’re just finding it explicitly.

Alternative Method — Factor First

Factor $f(x)$ before anything else:

f(x) = x^2 - 4x + 3 = (x - 1)(x - 3)

This immediately tells you the roots are $x = 1$ and $x = 3$ — which are exactly our interval endpoints. So $f(1) = 0$ and $f(3) = 0$ by inspection, without plugging in numbers.

For the vertex of $ax^2 + bx + c$ , the $x$ -coordinate is $x = -b/2a = 4/2 = 2$ . The vertex is the turning point, so $f'(2) = 0$ follows directly. Faster in an exam setting when you can recognise the parabola structure.

If the function is a quadratic $ax^2 + bx + c$ and the interval $[a, b]$ is given — always check if $a$ and $b$ are the roots. If yes, $c$ is always the midpoint $(a+b)/2$ by symmetry of parabolas. Here, $(1+3)/2 = 2$ . This saves about 30 seconds in JEE Main.

Common Mistake

Students write “f is continuous and differentiable — done” without actually verifying $f(1) = f(b)$ . In the CBSE marking scheme, checking all three conditions is mandatory. If you skip the endpoint check, you lose 1 mark even if your $c$ value is correct. Always compute $f(a)$ and $f(b)$ explicitly and state they are equal.

A subtler mistake: concluding Rolle’s Theorem doesn’t apply because the function has no turning point “visible”. Here $f(x) = x^2 - 4x + 3$ looks like it goes to zero at both ends, but students sometimes compute $f'(x) = 2x - 4$ , get $c = 2$ , and then forget to confirm $c \in (1, 3)$ . Always verify $c$ lies strictly inside the open interval — not at the endpoints.

Question

Solution — Step by Step

Why This Works

Alternative Method — Factor First

Common Mistake

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