
Rolle’s Theorem is a special case of the Mean Value Theorem (MVT).
It guarantees that if a continuous and differentiable function starts and ends at the same height, there must be at least one point between them where the slope of the tangent line is zero (that is, $$f^\prime(c) = 0$$.
Let $$f(x)$$ be a function such that:
$$f(x)$$ is continuous on the closed interval [a,b],
$$f(x)$$ is differentiable on the open interval (a,b),
$$f(a) = f(b)$$.
Then there exists at least one number c∈(a,b) such that:
$$f^\prime(c) = 0$$
Rolle’s Theorem says:
If you draw a smooth curve that begins and ends at the same height, the curve must have at least one horizontal tangent line somewhere in between.
That means the function must reach a peak or a valley (or flatten out) between a and b.
Imagine the curve passing through A(a,f(a)) and B(b,f(b)) at the same height.
Then, according to Rolle’s Theorem:
The tangent at some point C(c,f(c)) between A and B is horizontal.
So, $$f^\prime(c) = 0$$.
Let $$f(x) = x^2-4x+3$$ on the interval [1,3].
Step 1. Check conditions:
Continuous on [1,3]: ✅ (polynomial)
Differentiable on (1,3): ✅
$$f(1)=0 , f(3)=0 \longrightarrow f(1)=f(3)$$: ✅
Step 2. Compute $$f^\prime(x)$$ :
$$f^\prime(x) = 2x – 4$$
Step 3. Find $$f^\prime(c) = 0$$:
&&f^\prime(c) = 2c – 4 = 0 \longrightarrow c = 2&&
✅ Therefore, c=2 satisfies Rolle’s Theorem.
Let $$f(x)=\sin(x)$$ on $$[0 , \pi]$$.
Step 1. Check conditions:
Continuous and differentiable: ✅
$$f(0) = 0 , f(\pi) = 0 \longrightarrow f(0) = f(\pi)$$: ✅
Step 2. Compute $$f^\prime(x)$$:
$$f^\prime(x) = \cos(x)$$
Step 3. Find $$f^\prime(c) = 0$$:
$$f^\prime(c) = \cos(c) = 0\longrightarrow c = \frac{\pi}{2}$$
✅ So $$f^\prime(c)=0 at c=\frac{\pi}{2}$$
| Condition | Meaning |
|---|---|
| Continuous | No jumps or gaps in the curve |
| Differentiable | No sharp corners or cusps |
| $$f(a) = f(b)$$ | Function starts and ends at the same height |
| $$f^\prime(c)=0$$ | Tangent line is horizontal (flat) somewhere between a and b |
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