# Rolle's Theorem| Class 12 Maths

## Rolle's Theorem

If a function $f(x)$ satisfies the following three conditions:

1. $f(x)$ is continuous on the closed interval $[a, b]$.
2. $f(x)$ is differentiable on the open interval $(a, b)$.
3. $f(a) = f(b)$, meaning the function has the same values at both endpoints of the interval
Then there exists at least one $c$ in the open interval $(a, b)$ such that $f'(c) = 0$.
In other words, if a function is continuous on a closed interval, differentiable on the open interval, and has the same values at both endpoints of that interval, then there is at least one point $c$ inside the interval where the derivative of the function is zero. Geometrically, this means that the function has a horizontal tangent line at that point.

## Key points of Rolle's Theorem

1. Rolle's Theorem is a special case of the Mean Value Theorem (MVT), where the slope of the tangent line is guaranteed to be zero.
2. The conditions of continuity and differentiability are crucial for the theorem to apply. If the function is not continuous on the closed interval or not differentiable on the open interval, the theorem does not hold.
3. The theorem is often used to prove that if a function has the same values at its endpoints and is continuous and differentiable in between, there must be a point within the interval where the derivative is zero.
4. Rolle's Theorem is often used as a stepping stone to prove other important theorems in calculus, such as the Intermediate Value Theorem and the First Mean Value Theorem for Integrals.
5. While Rolle's Theorem guarantees the existence of at least one point $c$ with $f'(c) = 0$, it does not provide any information about how many such points there might be. There could be multiple points where the derivative is zero.

## Examples

Question 1
Consider the function $f(x) = x^3 - 6x^2 + 9x + 2$ on the closed interval $[1, 4]$.
a. Verify whether the conditions of Rolle's Theorem are satisfied for this function on the interval $[1, 4]$. State your reasons for each condition.
b. If the conditions of Rolle's Theorem are satisfied, find at least one point $c$ in the open interval $(1, 4)$ where $f'(c) = 0$. If they are not satisfied, explain why.
Solution
a. Conditions of Rolle's Theorem:
- Continuity on $[1, 4]$:The function $f(x) = x^3 - 6x^2 + 9x + 2$ is a polynomial function, and polynomial functions are continuous everywhere. So, it is continuous on $[1, 4]$.
- Differentiability on $(1, 4)$:The function is a polynomial, so it is differentiable everywhere. Thus, it is differentiable on $(1, 4)$.
- $f(1) = f(4)$: $f(1) = 1^3 - 6(1)^2 + 9(1) + 2 = 1 - 6 + 9 + 2 = 6$, and $f(4) = 4^3 - 6(4)^2 + 9(4) + 2 = 64 - 96 + 36 + 2 = 6$. So, $f(1) = f(4)$.

Therefore, all three conditions of Rolle's Theorem are satisfied for this function on $[1, 4]$.
b. Since the conditions of Rolle's Theorem are satisfied, we can conclude that there exists at least one point $c$ in the open interval $(1, 4)$ such that $f'(c) = 0$.
To find this point, we need to find the derivative of $f(x)$ and set it equal to zero:
$f(x) = x^3 - 6x^2 + 9x + 2$
$f'(x) = 3x^2 - 12x + 9$
Now, we set $f'(x) = 0$ and solve for $x$:
$3x^2 - 12x + 9 = 0$
Dividing the equation by 3:
$x^2 - 4x + 3 = 0$
Factoring:
$(x - 3)(x - 1) = 0$
So, $x = 1$ or $x = 3$.
Therefore, there are at least two points in the open interval $(1, 4)$ where $f'(c) = 0$: $c = 1$ and $c = 3$.

These are the answers to the questions based on Rolle's Theorem for the given function on the interval $[1, 4]$.
This is clear from the graph of the function

Question 2
Consider the function $f(x) = x^2 + 2$ on the closed interval $[-2, 2]$.
a. Verify whether the conditions of Rolle's Theorem are satisfied for this function on the interval $[-2, 2]$.. State your reasons for each condition.
b. If the conditions of Rolle's Theorem are satisfied, find at least one point $c$ in the open interval $(-2, 2)$ where $f'(c) = 0$. If they are not satisfied, explain why.
Solution
a. Conditions of Rolle's Theorem:
- Continuity on $[-2, 2]$: The function $f(x) = x^2 + 2$ is a polynomial function, and polynomial functions are continuous everywhere. So, it is continuous on$[-2, 2]$ .
- Differentiability on $(-2, 2)$:The function is a polynomial, so it is differentiable everywhere. Thus, it is differentiable on $(-2, 2)$.
- $f(-2) = f(2)$: $f(-2) = 6$, and $f(2) = 6$. So, $f(-2) = f(2)$.

Therefore, all three conditions of Rolle's Theorem are satisfied for this function on $[-2, 2]$.

b. Since the conditions of Rolle's Theorem are satisfied, we can conclude that there exists at least one point $c$ in the open interval $(-2, 2)$. such that $f'(c) = 0$.
To find this point, we need to find the derivative of $f(x)$ and set it equal to zero:
$f'(x) = 2x$
Now, we set $f'(x) = 0$ and solve for $x$:
$2x = 0$

So, $x = 0$

Thus at c = 0, we have f′(x) = 0

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