What is Cauchy’s Mean Value Theorem?
Cauchy’s Mean Value Theorem is a very important theorem used in the world of derivatives. The formula used in Cauchy's Theorem is
$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
It is very important to note that the given function should be continuous and differentiable over its entire domain. We will see this theorem in more detail in this article.
History of the Mathematician Cauchy
Augustin Louis Cauchy
Name: Augustin Louis Cauchy
Born: 21 August 1789
Died: 23 May 1857
Field: Mathematics
Nationality: French
General Statement of Cauchy’s Theorem
Cauchy’s Mean Value Theorem states that for any two functions $f\left( x \right)$ and$g\left( x \right)$, which are continuous on the interval $\left[ a,b \right]$ and differentiable on the interval $\left( a,\text{ }b \right)$ and $g\left( x \right)\ne 0$ for all $x\in \left( a,b \right)$, then there will be a point $x=c$ in the given interval such that
$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$.
Cauchy’s Theorem Proof
Let $f\left( x \right)$ and $g(x)$ be continuous on the interval $\left[ a,b \right]$ and differentiable on the interval $(a,b)$ and $g\left( x \right)\ne 0 \forall x\in \left( a,b \right)$.
Now let $F\left( x \right)=f\left( x \right)+P\times g\left( x \right)$
Where P is taken such that $F\left( x \right)$ satisfies the Rolle’s Theorem in $\left[ a,b \right]$
So, from Rolle’s Theorem $F\left( a \right)=F\left( b \right)$
$\Rightarrow f\left( a \right)+P\times g\left( a \right)=f\left( b \right)+P\times g\left( b \right)$
$\Rightarrow f\left( a \right)-f\left( b \right)=P\times\left( g\left( b \right)-g\left( a \right) \right)$
$\Rightarrow P=\dfrac{f(b)-f(a)}{g(b)-g(a)}$
Now $F(x)=f(x)-\left(\dfrac{f(b)-f(a)}{g(b)-g(a)}\right)g(x)$
Since $F\left( x \right)$ satisfies Rolle’s Theorem, $F\left( c \right)=0$
$\Rightarrow f'(c)-\left(\dfrac{f(b)-f(a)}{g(b)-g(a)}\right)g'(c)=0$
$\Rightarrow \dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
This proves the Cauchy Formula.
Hence proved.
Limitations of the Theorem
The first limitation of Cauchy's mean value theorem is that it is not applicable if the functions are not continuous in the given interval.
It is not applicable even if the given functions are not differentiable.
Applications of the Theorem
We know that from Cauchy’s Mean Value Theorem,
$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
Here if$g\left( x \right)=x$, the formula will be reduced to Lagrange’s Mean Value Theorem.
That is, $f'(c)=\dfrac{f(b)-f(a)}{b-a}$
Solved Examples
1. Find C of Cauchy’s Mean Value Theorem for the functions $\dfrac{1}{x}$ and $\dfrac{1}{{{x}^{2}}} \in \left[ 4,6 \right]$.
Ans. Let $f(x)=\dfrac{1}{x}$ and $g(x)=\dfrac{1}{{{x}^{2}}}$
Both $f\left( x \right)$and $g(x)$ are continuous and differentiable in $\left[ 4,6 \right]$
$g'(x)=\dfrac{-2}{{{x}^{3}}}$
So $g\left( x \right)\ne 0 \in \left[ 4,6 \right]$
Now we know that Cauchy’s Equation is
$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
$f'(x)=\dfrac{-1}{{{x}^{2}}}$
Now:
$f(a)=f(4)=\dfrac{1}{4}$
$f(b)=f(6)=\dfrac{1}{6}$
$g(a)=g(4)=\dfrac{1}{16}$
$g(b)=g(6)=\dfrac{1}{36}$
$f'(c)=-\dfrac{1}{{{c}^{2}}}$
$g'(c)=\dfrac{-2}{{{c}^{3}}}$
Substituting these values in the formula: $\dfrac{\dfrac{-1}{{{c}^{2}}}}{\dfrac{-2}{{{c}^{3}}}}=\dfrac{\dfrac{1}{6}-\dfrac{1}{4}}{\dfrac{1}{36}-\dfrac{1}{16}}$
By solving this, we get $c=4.8$.
2. If $f(x)$ is differentiable and $g\left( x \right)\ne 0$ such that $f(1)=4$,$f(2)=16$, $f\left( x \right)=8g\left( x \right)$ and $g(2)=4$ then what is the value of $g(1)$?
Ans: We know that from Cauchy’s Mean Value Theorem,
$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
Here $a=1$ and $b=2$
$\dfrac{f'(c)}{g'(c)}=\dfrac{f(2)-f(1)}{g(2)-g(1)}$
Now substituting all the given values in the equation:
$\dfrac{8g'(c)}{g'(c)}=\dfrac{16-4}{4-g(1)}$
$32-8g(1)=12$
$g(1)=\dfrac{5}{2}$
3. Using Cauchy’s Mean Value Theorem, find the mean value c for the function in $\left[ 2,5 \right]$ if $f\left( x \right)=\left| x-1 \right|$ and $g\left( x \right)=lnx$
Ans: $f\left( x \right)$ and $g\left( x \right)$are continuous and differentiable in the given interval.
We know that from Cauchy’s Mean Value Theorem,
$\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
So, $f'(x)=\dfrac{\left| x-1 \right|}{x-1}$ and $g'(x)=\dfrac{1}{x}$
Now $f'(c)=1$ and $g'(c)=\dfrac{1}{c}$
$f\left( a \right)=f(2)=\left| 2-1 \right|=1$
$f\left( b \right)=f(5)=\left| 5-1 \right|=4$
Now substituting these values in the formula:
$\dfrac{1}{\dfrac{1}{c}}=\dfrac{4-1}{\ln 5-\ln 2}$
$c=\dfrac{3}{\ln \left(\dfrac{5}{2}\right)}=3.27$
Conclusion
In conclusion, we learn that Cauchy’s Mean Value Theorem is derived with the help of Rolle’s Theorem. Lagrange’s mean value theorem can be deduced from Cauchy’s Mean Value Theorem. Cauchy’s Mean Value Theorem is the relationship between the derivatives of two functions and changes in these functions on a finite interval. The continuity and differentiability of the given functions must be adequately checked.
Important Points to Rememeber
The continuity of the given functions must be checked.
The differentiability of the given functions must also be checked.
The formula must be applied properly.
Important Formulae:
The final formula of Cauchy’s Mean value theorem is $\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{f'(c)}{g'(c)}$
The final formula of Rolle’s theorem is $F\left( a \right)=F\left( b \right)$
FAQs on Cauchy’s Mean Value Theorem
1. What is Cauchy’s Mean Value Theorem (CMVT)?
Cauchy’s Mean Value Theorem is an extension of Lagrange’s Mean Value Theorem. It states that if two functions, f(x) and g(x), are continuous on a closed interval [a, b] and differentiable on the open interval (a, b), with g'(x) ≠ 0 for any x in (a, b), then there exists at least one point 'c' in (a, b) such that: (f(b) - f(a)) / (g(b) - g(a)) = f'(c) / g'(c). This formula establishes a relationship between the ratio of the change in the two functions over the interval and the ratio of their derivatives at a specific point 'c' within that interval.
2. What is the importance of the conditions required for Cauchy’s Mean Value Theorem?
The conditions are essential because they guarantee the theorem's conclusion. Here’s why each is important:
- Continuity on [a, b]: This ensures there are no breaks, jumps, or holes in the functions' graphs, which is necessary for the average rate of change to be well-defined across the entire interval.
- Differentiability on (a, b): This ensures that both functions have a defined tangent (and thus a defined instantaneous rate of change) at every point inside the interval. Without this, f'(c) or g'(c) might not exist.
- g'(x) ≠ 0 on (a, b): This is a critical condition to prevent division by zero in the formula. If g'(c) were zero, the ratio f'(c)/g'(c) would be undefined. It also implies, by Rolle's Theorem, that g(a) ≠ g(b), ensuring the denominator on the other side is also not zero.
3. How does Cauchy’s Mean Value Theorem relate to Lagrange’s Mean Value Theorem?
Lagrange’s Mean Value Theorem (LMVT) is a special case of Cauchy's Mean Value Theorem. You can derive LMVT from CMVT by setting the function g(x) = x. If you substitute g(x) = x into the CMVT formula, you get:
- g(b) = b and g(a) = a
- g'(x) = 1, so g'(c) = 1
The CMVT formula, (f(b) - f(a)) / (g(b) - g(a)) = f'(c) / g'(c), then simplifies to (f(b) - f(a)) / (b - a) = f'(c) / 1, which is the exact formula for Lagrange's Mean Value Theorem.
4. Can you explain the geometrical interpretation of Cauchy's Mean Value Theorem?
Geometrically, Cauchy's Mean Value Theorem can be understood by considering a parametric curve defined by x = g(t) and y = f(t), where t is in the interval [a, b]. The theorem states that there is at least one point 'c' between 'a' and 'b' on the curve where the tangent line is parallel to the secant line connecting the endpoints of the curve, which are (g(a), f(a)) and (g(b), f(b)). The slope of the secant line is (f(b) - f(a)) / (g(b) - g(a)), and the slope of the tangent at t = c is f'(c) / g'(c).
5. Why must a function be differentiable on the open interval (a, b) and not necessarily the closed interval [a, b] for the theorem to apply?
This is a fundamental concept in calculus. For a function to be differentiable at a point, its derivative must exist, which requires the limit defining the derivative to be the same from both the left and the right. At the endpoints of a closed interval [a, b]:
- At point 'a', we can only approach it from the right side (within the interval).
- At point 'b', we can only approach it from the left side.
Since we cannot check for the limit from both sides at the endpoints, a function is generally not considered differentiable at the exact endpoints of its domain. The theorem only requires differentiability within the interval, hence the open interval (a, b).
6. How do you find the value 'c' in an example using Cauchy's Mean Value Theorem?
To find the value 'c', you follow these steps. For example, for functions f(x) = 1/x and g(x) = 1/x² in the interval [4, 6]:
- Verify conditions: Confirm f(x) and g(x) are continuous and differentiable on [4, 6] and g'(x) ≠ 0.
- Find derivatives: f'(x) = -1/x² and g'(x) = -2/x³.
- Calculate function values at endpoints: f(4)=1/4, f(6)=1/6, g(4)=1/16, g(6)=1/36.
- Set up the CMVT equation:
(f(6) - f(4)) / (g(6) - g(4)) = f'(c) / g'(c)
((1/6) - (1/4)) / ((1/36) - (1/16)) = (-1/c²) / (-2/c³) - Solve for c: Simplifying the equation gives c/2 = 12/5, which leads to c = 24/5 or 4.8. Since 4.8 is within the interval (4, 6), this is the correct value.
7. When does Cauchy's Mean Value Theorem fail to apply?
Cauchy's Mean Value Theorem cannot be applied if any of its core conditions are not met. The theorem fails if:
- Either f(x) or g(x) is not continuous on the closed interval [a, b].
- Either f(x) or g(x) is not differentiable on the open interval (a, b).
- The derivative g'(x) is equal to zero for any point x within the interval (a, b).
If any of these conditions are violated, there is no guarantee that a point 'c' satisfying the theorem's equation exists.
8. Is Cauchy's Mean Value Theorem also known by another name?
Yes, Cauchy's Mean Value Theorem is sometimes referred to as the Extended Mean Value Theorem or the Second Mean Value Theorem. This name highlights its role as a generalization of the standard Mean Value Theorem (Lagrange's Mean Value Theorem).
