Rolle’s Theorem and Lagrange’s Mean Value Theorem

Introduction

Rolle’s theorem and Lagrange’s mean value theorem are interpreted on a function over an interval if the function satisfies the condition of continuity over a given closed interval and the condition of differentiability over a given open interval. The continuity of a function over a closed interval is defined as the function's graph that should not contain any break over the interval. The differentiability of a function over an open interval is defined as the function should be differentiable at every point in the interval.

Continuity and Differentiability

Continuity: Let’s take a function 𝑓(𝑥) with a domain and a point 𝑘 which is in the domain of the function.

The function 𝑓(𝑥) at point 𝑘 is continuous if $\mathrm{\lim_{x\:\rightarrow\:k}f(x)\:=\:f(k)}$

Precisely, at $\mathrm{x\:=\:k:\:\lim_{x\:\rightarrow\:k}f(x)\:=\:f(k)\:=\:\lim_{x\:\rightarrow\:k}f(x)}$

The function is continuous in a closed interval [𝑎, 𝑏] if it satisfies the following conditions.

• The function should be continuous over its open interval (𝑎, 𝑏)

• $\mathrm{\lim_{x\:\rightarrow\:a}f(x)\:=\:f(a)}$

• $\mathrm{\lim_{x\rightarrow\:b}f(x)\:=\:f(b)}$

Differentiability: Assume a function 𝑓(𝑥) its derivative is 𝑓′(𝑥) if the limit exists

then $\mathrm{f'(x)\:=\:\lim_{h\rightarrow\:0}\:\frac{f(x\:+\:h)\:-\:f(x)}{h}}$ The function is differentiable at a point if $\mathrm{\lim_{h\rightarrow\:0}\:\frac{f(k\:+\:h)\:-\:f(k)}{h}\:=\:\lim_{h\rightarrow\:0}\frac{f(k\:+\:h)\:-\:f(k)}{h}}$ of the function at x = k is equal and finite.

In an open interval, the function is differentiable if it is differentiable at every point of the interval.

In a closed interval [𝑎, 𝑏] the function is differentiable if it satisfies the following conditions.

• If the right-hand derivative at 𝑎 for the function $\mathrm{\lim_{h\rightarrow\:0^{-}}\frac{f(a\:+\:h)\:-\:f(a)}{h}}$ and the left-hand derivative at 𝑏 for the function $\mathrm{\lim_{h\rightarrow\:0^{-}}\frac{f(b\:+\:h)\:-\:f(b)}{h}}$

• $\mathrm{f'(x)}$ of the function exists at every point of the interval (𝑎, 𝑏)

Rolle's Theorem

Rolle’s theorem is defined as if a function defined over [𝑎, 𝑏] satisfies the following conditions−

• The function is continuous over [𝑎, 𝑏]

• The function is differentiable over (𝑎, 𝑏)

then there exists at least one point let’s say 𝑘 in (𝑎, 𝑏) such that $\mathrm{f'(k)\:=\:0}$

There can be more than one value but one value exists.

Rolle’s theorem is also defined as if $\mathrm{f(a)\:=\:f(b)}$ then there exists at least one point let’s say 𝑘 in (𝑎, 𝑏) such that $\mathrm{f'(k)\:=\:0}$

If a polynomial function $\mathrm{f(x)\:=\:0}$ has two roots then there exists at least one root for $\mathrm{f'(x)\:=\:0}$ between the roots of the polynomial function $\mathrm{f(x)\:=\:0}$

Lagrange's Mean Value Theorem

It is defined as if a function defined over an interval [𝑎, 𝑏] satisfies the following conditions −

• The function is continuous over [𝑎, 𝑏]

• The function is differentiable over (𝑎, 𝑏)

then there exists at least one point let’s say 𝑘 in (𝑎, 𝑏) such that $\mathrm{f'(k)\:=\:\frac{f(b)\:-\:f(a)}{b\:-\:a}}$ If $\mathrm{f(b)\:=\:f(a)\:then\:f'(k)\:=\:0}$ is Rolle’s theorem

Intermediate Value Theorem

In general, words assume a continuous graph of a function take two points in the continuous graph and consider a line such that the two points are on opposite sides of the line then it should intersect the graph at least once of course, it must cross the line for the two points on the opposite side this is the basic idea of intermediate value theorem

Definition: A function $\mathrm{f(x)}$ which is continuous over an interval [a, b] where 𝑘 is a value between 𝑓(𝑎) and 𝑓(𝑏) then there must be at least one value let’s say 𝑐 between [a, b] such that $\mathrm{f(c)\:=\:k}$. The definition says at least one value but there is a possibility of more than one value.

Solved Example

1) Check whether the following function satisfies Rolle’s theorem $\mathrm{f(x)\:=\:x^{2}\:+\:1\:for\:[-3\:,\:3]}$ ?

The given function $\mathrm{f(x)\:=\:x^{2}\:+\:1}$ continuous and differentiable over the interval.

$$\mathrm{f(x)\:=\:x^{2}\:+\:1}$$

$\mathrm{f(3)\:=\:x^{2}\:+\:1\:=\:9\:+\:1\:=\:10}$

$\mathrm{f(-3)\:=\:x^{2}\:+\:1\:=\:9\:+\:1\:=\:10}$

$$\mathrm{f(3)\:=\:f(-3)\:10}$$

from the definition of Rolle’s theorem, we know that let’s say 𝑘 in (-3, 3) such that $\mathrm{f'(k)\:=\:0}$

$$\mathrm{f(x)\:=\:x^{2}\:+\:1}$$

$\mathrm{f'(x)\:=\:0}$

$$\mathrm{x\:=\:0}$$

0 lies between (−3, 3). Hence proved.

2) Check whether the following function satisfies Lagrange’s theorem $\mathrm{f(x)\:=\:x^{2}\:+\:2x\:+\:1\:for\:[-2\:,\:2]}$

The given function $\mathrm{f(x)\:=\:x^{2}\:+\:2x\:+\:1}$ defined over an interval $\mathrm{[-2\:,\:2]}$

$\mathrm{f(x)\:=\:x^{2}\:+\:2x\:+\:1}$ is continuous over the interval $\mathrm{[-2\:,\:2]}$

$\mathrm{f(x)\:=\:x^{2}\:+\:2x\:+\:1}$ is differentiable over the interval $\mathrm{(-2\:,\:2)}$

$$\mathrm{f(x)\:=\:x^{2}\:+\:2x\:+\:1}$$

$$\mathrm{f'(x)\:=\:2x\:+\:2}$$

$\mathrm{f(-2)\:=\:x^{2}\:+\:2x\:+\:1\:=\:4\:-\:4\:+\:1\:=\:1}$

$\mathrm{f(2)\:=\:x^{2}\:+\:2x\:+\:1\:=\:4\:+\:4\:+\:1\:=\:9}$

from Lagrange’s theorem, we know that $\mathrm{f'(k)\:=\:\frac{f(b)\:-\:f(a)}{b\:-\:a}\:=\:2}$

$$\mathrm{f'(k)\:=\:2}$$

$$\mathrm{f'(k)\:=\:2k\:+\:2\:=\:2}$$

𝑘 = 0 lies between [−2, 2]. Hence Proved.

Conclusion

In this tutorial, we learned about continuity and differentiability of functions over an interval, Rolle’s theorem for a function at an interval, Lagrange’s theorem for a function at an interval, and an Intermediate value theorem for a function at an interval, and a few examples.

FAQs

1. Define Rolle’s theorem and Lagrange’s mean value theorem?

Rolle’s theorem states that there exists at least one point let’s say 𝑘 in (𝑎, 𝑏) such that $\mathrm{f'(k)\:=\:0}$ Lagrange’s theorem states that there exists at least one point let’s say 𝑘 in (𝑎, 𝑏) such that $\mathrm{f'(k)\:=\:\frac{f(b)\:-\:f(a)}{b\:-\:a}}$

2. How to verify that a function is continuous over a closed interval?

In a closed interval [𝑎, 𝑏] the function is continuous

• if the function is continuous over its open interval (𝑎, 𝑏)

• $\mathrm{\lim_{x\rightarrow\:a}f(x)\:=\:f(a)}$

• $\mathrm{\lim_{x\rightarrow\:b}f(x)\:=\:f(b)}$

3. How to verify that a function is differentiable over a closed interval?

s. In a closed interval [𝑎, 𝑏] the function is differentiable if the right-hand derivative at 𝑎 for the function $\mathrm{\lim_{h\rightarrow\:0}\frac{f(a\:+\:h)\:-\:f(a)}{h}}$ exist, the left-hand derivative at 𝑏 for the function $\mathrm{\lim_{h\rightarrow\:0}\frac{f(b\:+\:h)\:-\:f(b)}{h}}$ exist, and $\mathrm{f'(x)}$ of the function exists at every point of the interval (𝑎, 𝑏).

4. Check the polynomial function $\mathrm{f(x)\:=\:x^{2}\:+\:x\:-\:6}$ has roots between [𝟎, 𝟑] using the intermediate value theorem?

Given $\mathrm{f(x)\:=\:x^{2}\:+\:x\:-\:6}$ is continuous in [0, 3] because it is a polynomial function.

At $\mathrm{x\:=\:0\:\colon\:f(x)\:=\:x^{2}\:+\:x\:-\:6\:=\:-6}$

At $\mathrm{x\:=\:3\:\colon\:f(x)\:=\:x^{2}\:+\:x\:-\:6\:=\:6}$

To have a root the graph of the polynomial function must pass through y = 0. The values of the function at 0 and 3 are on opposite sides of the line y = 0, or the number 0 is between 𝑓(0) and 𝑓(3). Therefore, the graph must pass through y = 0 in [0, 3]. It has a root in between [0, 3].

5. Check the polynomial function $\mathrm{f(x)\:=\:x^{2}\:-\:5x\:+\:6}$ has roots between [-3, 0] using the intermediate value theorem?

Given $\mathrm{f(x)\:=\:x^{2}\:-\:5x\:+\:6}$ is continuous in [-3, 0] because it is a polynomial function

At $\mathrm{x\:=\:-3\:\colon\:f(x)\:=\:x^{2}\:-\:-\:5x\:+\:6}$

At $\mathrm{x\:=\:0\:\colon\:f(x)\:=\:x^{2}\:-\:5x\:+\:6\:=\:6}$

To have a root the graph of the polynomial function must pass through y = 0.

The values of the function at -3 and 0 are on the same side of the line y = 0, or the number 0 is not between 𝑓(0) and 𝑓(−3).

Therefore, the graph won’t pass through y = 0 in [-3, 0]. It doesn’t have a root in between [-3, 0].