Cauchy's Mean Value Theorem
Let \( f(x) \) and \( g(x) \) be two functions that are continuous on the closed interval \([a,b] \) and differentiable on the open interval \((a,b) \). Assume that \( g'(x)\neq0 \) for every \( x\in(a,b) \). Then there exists at least one point \( c\in(a,b) \) such that
\[ \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)} \]
In other words, the ratio of the overall changes in the two functions across the interval is equal to the ratio of their instantaneous rates of change at some point inside the interval.
Cauchy's Mean Value Theorem establishes a fundamental connection between the average behavior of two functions and their local behavior. Whenever the required regularity conditions are satisfied, there is at least one interior point where the ratio of the derivatives is exactly the same as the ratio of the functions' overall changes.
Why is this theorem important? Cauchy's Mean Value Theorem generalizes the Lagrange Mean Value Theorem to pairs of functions. If we choose \( g(x)=x \), then \( g'(x)=1 \), and the identity \[ \frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f'(c)}{g'(c)} \] reduces to \[ \frac{f(b)-f(a)}{b-a}=f'(c) \] which is precisely the statement of the Lagrange Mean Value Theorem. Because of this, Cauchy's theorem is regarded as a natural extension of Lagrange's result. It also plays a central role in the proof of l'Hôpital's rule, making it one of the key results in differential calculus.
A worked example
Consider the functions \( f(x)=x^2 \) and \( g(x)=x+1 \) on the interval \( [1,3] \).
Both functions are continuous and differentiable, and
\[ g'(x)=1\neq0 \]
First, compute the ratio of the functions' increments:
\[ \frac{f(3)-f(1)}{g(3)-g(1)} = \frac{9-1}{4-2} = 4 \]
The derivatives are
\[ f'(x)=2x \qquad g'(x)=1 \]
Therefore, the ratio of the derivatives is
\[ \frac{f'(c)}{g'(c)} = 2c \]
Setting this equal to the ratio of the increments gives
\[ 2c=4 \]
so that
\[ c=2 \]
Since \( c=2 \) lies in the interval \( (1,3) \), all the hypotheses are satisfied and the conclusion of Cauchy's Mean Value Theorem holds.
Geometric interpretation
At the point \( c=2 \), the ratio of the instantaneous rates of change of the two functions is exactly the same as the ratio of their total changes over the interval \([1,3]\).
Equivalently, \( c=2 \) is the point where the ratio of the slopes of the tangent lines to the graphs of \( f \) and \( g \) matches the ratio of the slopes of the secant lines joining the endpoints of the interval.
Note. The condition \( g'(x)\neq0 \) is essential. It guarantees that the ratio of the derivatives is always well defined by ensuring that the denominator never vanishes.
Proof
Assume that \( f(x) \) and \( g(x) \) are continuous on \( [a,b] \) and differentiable on \( (a,b) \).
To prove Cauchy's Mean Value Theorem, define the auxiliary function
\[ F(x)=f(x)-kg(x) \]
where \( k\in\mathbb{R} \).
Since both \( f(x) \) and \( g(x) \) satisfy the required continuity and differentiability conditions, the function \( F(x) \) is also continuous on \([a,b]\) and differentiable on \((a,b)\).
Next, choose the constant \( k \) so that
\[ F(a)=F(b) \]
This requires
\[ f(a)-kg(a)=f(b)-kg(b) \]
Collecting the terms containing \( k \) gives
\[ k[g(b)-g(a)]=f(b)-f(a) \]
Hence,
\[ k=\frac{f(b)-f(a)}{g(b)-g(a)} \]
Substituting this value into the auxiliary function yields
\[ F(x)=f(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g(x) \]
With this choice of \( k \), the function \( F(x) \) satisfies the hypotheses of Rolle's Theorem.
Therefore, there exists at least one point \( c\in(a,b) \) such that
\[ F'(c)=0 \]
Differentiate \( F(x) \):
\[ F'(x)=f'(x)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(x) \]
Evaluating the derivative at the point \( c \) gives
\[ F'(c)=f'(c)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)=0 \]
Therefore,
\[ f'(c)-\frac{f(b)-f(a)}{g(b)-g(a)}g'(c)=0 \]
which implies
\[ f'(c)=\frac{f(b)-f(a)}{g(b)-g(a)}g'(c) \]
Finally, divide both sides by \( g'(c) \). Since \( g'(c)\neq0 \), this operation is valid:
\[ \frac{f'(c)}{g'(c)}=\frac{f(b)-f(a)}{g(b)-g(a)} \]
Thus, there exists at least one point inside the interval \( (a,b) \) where the ratio of the derivatives of the two functions is equal to the ratio of their increments.
\[ \frac{f(b)-f(a)}{g(b)-g(a)}=\frac{f'(c)}{g'(c)} \]
This completes the proof of Cauchy's Mean Value Theorem.
And so on.
