Lagrange’s mean value theorem states that if a function $ƒ(x)$ is

- continuous in the closed interval $[a,b]$
- differential in the open interval $(a,b)$

then there exists at least one point $c\in(a,b)$ such that \[ƒ'(c)=\frac{ƒ(b)-ƒ(a)}{b-a}\]

## Geometrical Interpretation of Mean Value Theorem

Geometrically, Lagrange’s mean value theorem says that in a continuous curve [Continuity], in which tangent can be drawn at every point, there is at least one point where the tangent is parallel to the secant joining the end points as shown in the following figure.

Putting $b=a+h$ so that $b-a=h$ is the length of the interval, the form of Lagrange’s mean value theorem can be put in the form \[ƒ(a+h)=ƒ(a)+hƒ'(a+\theta h)\] where $\theta$ is a positive number less than $1$ i.e. $0<\theta<1$.

### Verify Lagrange’s mean value theorem for the function $ƒ(x)=x^2-2x+4$ in $[1,5]$.

Here, \[ƒ(x)=x^2-2x+4\]

- $ƒ(x)$ is polynomial, so it is continuous in $[1,5]$.
- $ƒ'(x)=2x-2$, which exists for all $x\in(1,5)$. So, $ƒ(x)$ is differentiable in $(1,5)$.

Hence, all the conditions of mean value theorem are satisfied. So, there exists at least one point $c\in(1,5)$ such that \[ƒ'(c)=\frac{ƒ(5)-ƒ(1)}{5-1}\text{ __(1)}\] Now, \[ƒ(5)=5^2-2×5+4=19\] \[ƒ(1)=1^2-2×1+4=3\] \[ƒ'(c)=2c-2\] Putting these values in $\text{(1)}$, \[2c-2=\frac{19-3}{5-1}\] \[2c-2=4\] \[\therefore c=3\in(1,5)\] Hence, mean value theorem is verified.

### Verify Lagrange’s mean value theorem for the function $ƒ(x)=x^3+x^2-6x$ in $[-1,4]$.

Here, \[ƒ(x)=x^3+x^2-6x\]

- $ƒ(x)$ is polynomial, so it is continuous in $[-1,4]$.
- $ƒ'(x)=3x^2+2x-6$, which exists for all $x\in(-1,4)$. So, $ƒ(x)$ is differentiable in $(-1,4)$.

Hence, all the conditions of mean value theorem are satisfied. So, there exists at least one point $c\in(-1,4)$ such that \[ƒ'(c)=\frac{ƒ(4)-ƒ(-1)}{4-(-1)}\text{ __(1)}\] Now, \[ƒ(-1)=(-1)^3+(-1)^2+6=6\] \[ƒ(4)=4^3+4^2-6×4=56\] \[ƒ'(c)=3c^2+2c-6\] Putting these values in $\text{(1)}$, \[3c^2+2c-6=\frac{56-6}{4-(-1)}\] \[3c^2+2c-16=0\] \[(c-1)(3c+8)=0\] \[\text{Either, }c=2\text{ or }c=\frac{-8}{3}\] \[\therefore c=2\in(-1,4) \text{ and } c=\frac{-8}{3}\in(-1,4)\] Hence, mean value theorem is verified.

### Verify Lagrange’s mean value theorem for the function $ƒ(x)=\sqrt{x^2-4}$, $x\in[2,4]$.

Here, \[ƒ(x)=\sqrt{x^2-4}\]

- $ƒ(x)=\sqrt{x^2-4}$ exists for all $x\in[2,4]$, so it is continuous in $[-1,4]$.
- $ƒ'(x)=\frac{x}{\sqrt{x^2-4}}$, which exists for all $x\in(2,4)$. So, $ƒ(x)$ is differentiable in $(2,4)$.

Hence, all the conditions of mean value theorem are satisfied. So, there exists at least one point $c\in(2,4)$ such that \[ƒ'(c)=\frac{ƒ(4)-ƒ(2)}{4-2}\text{ __(1)}\] Now, \[ƒ(4)=\sqrt{4^2-4}=2\sqrt{3}\] \[ƒ(2)=\sqrt{2^2-4}=0\] \[ƒ'(c)=\frac{c}{\sqrt{c^2-4}}\] Putting these values in $\text{(1)}$, \[\frac{c}{\sqrt{c^2-4}}=\frac{2\sqrt{3}-0}{4-2}\] \[\frac{c^2}{c^2-4}=3\] \[2c^2=12\] \[\therefore c=\sqrt{6}\in(2,4)\] Hence, mean value theorem is verified.

### Verify Lagrange’s mean value theorem for the function $ƒ=e^x$, $x\in[0,1]$.

Here, \[ƒ(x)=e^x\]

- $ƒ(x)=e^x$ exists for all $x\in[0,1]$, so it is continuous in $[0,1]$.
- $ƒ'(x)=e^x$, which exists for all $x\in(0,1)$. So, $ƒ(x)$ is differentiable in $(0,1)$.

Hence, all the conditions of mean value theorem are satisfied. So, there exists at least one point $c\in(0,1)$ such that \[ƒ'(c)=\frac{ƒ(1)-ƒ(0)}{1-0}\text{ __(1)}\] Now, \[ƒ(1)=e\] \[ƒ(0)=e^0=1\] \[ƒ'(c)=e^c\] Putting these values in $\text{(1)}$, \[e^c=\frac{e-1}{1-0}\] \[e^c=e-1\] \[\log e^c=\log(e-1)\] \[c\log e=\log(2.7182-1)\] \[c=\log(1.7182)\] \[\therefore c=0.23\in(0,1)\] Hence, mean value theorem is verified.

### Examine whether the function $ƒ(x)=x^2-6x+1$ satisfies Lagrange’s mean value theorem. If it satisfies, find the coordinates of the point at which the tangent is parallel to the chord joining the points $A(1,-4)$ and $B(3,-8)$.

Here, \[ƒ(x)=x^2-6x+1\text{ in }[1,3]\]

- $ƒ(x)$ is polynomial, so it is continuous in $[1,3]$.
- $ƒ'(x)=2x-6$, which exists for all $x\in(1,3)$. So, $ƒ(x)$ is differentiable in $(1,3)$.

Hence, all the conditions of mean value theorem are satisfied. So, there exists at least one point $c\in(1,3)$ such that \[ƒ'(c)=\frac{ƒ(3)-ƒ(1)}{3-1}\text{ __(1)}\] Now, \[ƒ(3)=3^2-6×3+1=-8\] \[ƒ(1)=1^2-6×1+1=-4\] \[ƒ'(c)=2c-6\] Putting these values in $\text{(1)}$, \[2c-6=\frac{-8-(-4)}{3-1}\] \[2c-6=-2\] \[\therefore c=2\in(1,3)\] Hence, mean value theorem is verified.

$c=2$ is the $\text{x-coordinate}$ of the point at which the tangent drawn is parallel to the chord joining the points $A(1,-4)$ and $B(3,-8)$. \[\therefore ƒ(2)=2^2-6×2+1=-7\] Therefore, the required point is $(2,-7)$.

### Using mean value theorem for the function $ƒ(x)=\sin x$ in $[0,x]$ $(0<x<\frac{π}{2})$ prove that $\sin x<x$.

Here, \[ƒ(x)=\sin x\]

- $ƒ(x)=\sin x$ exists for all $x\in[0,x]$, so it is continuous in $[0,x]$.
- $ƒ'(x)=\cos x$, which exists for all $x\in(0,x)$. So, $ƒ(x)$ is differentiable in $(0,x)$.

Hence, all the conditions of mean value theorem are satisfied. So, there exists at least one point $c\in(0,x)$ such that \[ƒ'(c)=\frac{ƒ(x)-ƒ(0)}{x-0}\] \[\Rightarrow \cos c=\frac{\sin x-0}{x-0}\] \[\Rightarrow \cos c=\frac{\sin x}{x}\] \[\Rightarrow \frac{\sin x}{x}<1\] \[\Rightarrow \sin x<x\]