Rolle’s Theorem - Practice Questions & MCQ

Rolle’s Theorem

Let $f$ be a real-valued function defined on the closed interval $[a, b]$ such that
1. it is continuous on the closed interval $[\mathrm{a}, \mathrm{b}]$,
2. it is differentiable on the open interval ( $a$, $b$ ), and
3. $f(a)=f(b)$.

There then exists at least one $c \in(a, b)$ such that $f^{\prime}(c)=0$.

Geometrical interpretation of Rolle's Theorem: are equal. Then there exists at least one point $(c, f(c))$ lying between $A$ and $B$ on the curve $y=f(x)$ where the tangent is parallel to the $x$-axis. I.e. $f^{\prime}(c)=0$.

roots of the equation $f^{\prime}(x)=0$, there exists at least one root of the equation $f^{\prime}(x)=0$ and so on.

Note:
 The converse of Rolle's theorem need not be true.

For example,
Let $f(x)=x^3-x^2-x+1$ in the interval $[-1,2]$

$
\begin{aligned}
f^{\prime}(x) & =3 x^2-2 x-1 \\
f^{\prime}(1) & =3(1)^2-2(1)-1=0
\end{aligned}
$

but $\mathrm{f}(-1) \neq \mathrm{f}(2)$

Application of Rolle's Theorem:
We can use Rolle's theorem to study and calculate the root of the equation as well as locate the roots, The equation must be continuous as well as differentiable.
Let's look at one example.
Consider the equation $f(x)=(x-1)(x-2)(x-3)(x-4)$. Then how many roots of the equation $f^{\prime}(x)=0$ are positive?
Let $f(x)=(x-1)(x-2)(x-3)(x-4)$.
Since, $f(x)$ is continuous and differentiable and $f(1)=f(2)=f(3)=f(4)=0$
According to Rolle's theorem, there is at least one root of the equation $f^{\prime}(x)=0$ in each of the intervals $(1,2),(2,3)$ and $(3,4)$.
Since, $f^{\prime}(x)$ is a cubic function. So $f^{\prime}(x)=0$ has exactly one root in each interval $(1,2),(2,3)$ and $(3,4)$

Lagrange’s Mean Value Theorem (LMVT)

Lagrange's Mean Value Theorem generalized Rolle’s theorem by considering functions that do not necessarily have equal value at the endpoints. 

Statement

Let $f(x)$ be a function defined on $[a, b]$ such that
1. it is continuous on [a, b],
2. it is differentiable on (a, b).

Then there exists at least one real number $c \in(a, b)$ such that.

$
f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}
$
Geometrical Interpretation: