Cauchy's Mean Value Theorem is an important part in proving l'Hospital's Rule and as such, it is important to have a basic understanding of the Theorem.[br][br]Proving Cauchy's Mean Value Theorem is very similar to proving the Mean Value Theorem and we will address this later in the activity.[br][br]Before that, we begin by introducing the theorem.

Suppose that the functions [math]f[/math] and [math]g[/math] are continuous on [math][a,b][/math] and differentiable on [math](a,b)[/math], and [math]g^{\prime}(x)\neq 0[/math] for all [math]x[/math] in [math](a,b)[/math]. Then there is a number [math]c[/math] in [math](a,b)[/math] such that[br][br][center][math]\dfrac{f^{\prime}(c)}{g^{\prime}(c)} = \dfrac{f(b) - f(a)}{g(b) - g(a)}[/math][/center][br]Now, before we prove the theorem, let us look at an example to build some intuition.

Let [math]f(x) = e^{x} - 1[/math] and [math]g(x) = x^3 + 4x[/math] and consider the interval [math][-1,1][/math].[br][br]First, we see that since [math]f(x) = e^{x} - 1[/math] and [math]g(x) = x^{3} + 4x[/math] are both continuous for all [math]x[/math], they are continuous on the interval [math][-1,1][/math].[br][br]Second, their derivatives [math]f^{\prime}(x) = e^{x}[/math] and [math]g^{\prime}(x) = 3x^{2} + 4[/math] are also continuous for all [math]x[/math], and are therefore continuous on the interval [math](-1,1)[/math].[br][br]Finally, we see that [math]g^{\prime}(x) > 0[/math] for all [math]x[/math], and thus clearly [math]g^{\prime}(x)\neq 0[/math] on [math](-1,1)[/math].[br][br]Thus, there exists a number [math]c[/math] in the interval [math](-1,1)[/math] such that[br][br][center][math]\dfrac{f^{\prime}(c)}{g^{\prime}(c)} = \dfrac{f(1) - f(-1)}{g(1) - g(-1)}[/math][/center][br]Replacing the values into the functions and simplifying, we have[br][br][center][math]\dfrac{e^{c}}{3c^{2} + 4} = \dfrac{e^{1} - 1 - e^{-1} + 1}{(1)^{3} + 4(1) - (-1)^{3} - 4(-1)} = \dfrac{e - e^{-1}}{10}[/math][/center][br]So, we have[br][br][center][math]\dfrac{e^{c}}{3c^{2} + 4} = \dfrac{e - e^{-1}}{10}[/math][/center][br]Since we cannot solve this for [math]c[/math] directly, we will try to estimate the value using the graphic below. [br][br]Use the slider to move the point closer to the point of intersection and then use the input box above to get a closer approximation.[br]

Suppose that [math]f(x)[/math] and [math]g(x)[/math] are functions that are continuous on [math][a,b][/math], differentiable on [math](a,b)[/math], and [math]g^{\prime}(x)\neq 0[/math] for all [math]x[/math] on [math](a,b)[/math].[br][br]First, we note that the denominator on the right-hand side is not zero. If [math]g(b) - g(a) = 0[/math] then [math]g(b) = g(a)[/math] and by Rolle's Theorem there exists a number [math]c[/math] in the interval [math](a,b)[/math] such that [math]g^{\prime}(c) = 0[/math]. This contradicts the hypothesis that [math]g^{\prime}(x)\neq 0[/math] for all [math]x[/math] in [math](a,b)[/math]. Thus, [math]g(b) - g(a) \neq 0[/math].[br][br]Next, we define a new function[br][br][center][math]h(x) = f(x) - f(a) - \dfrac{f(b) - f(a)}{g(b) - g(a)}\left[g(x) - g(a)\right][/math][/center][br]This new function is continuous on [math][a,b][/math], differentiable on [math](a,b)[/math].[br][br]Additionally, this new function satisfies the condition that [math]h(a) = h(b)[/math] since,[br][br][center][math]h(b) = f(b) - f(a) - \dfrac{f(b) - f(a)}{g(b) - g(a)}\left[g(b) - g(a)\right] = 0[/math][/center][br]and[br][br][center][math]h(a) = f(a) - f(a) - \dfrac{f(b) - f(a)}{g(b) - g(a)}\left[g(a) - g(a)\right] = 0[/math][/center][br]Therefore, by Rolle's Theorem there exists a number [math]c[/math] in [math](a,b)[/math] such that[br][br][center][math]h^{\prime}(c) = f^{\prime}(c) - \dfrac{f(b) - f(a)}{g(b) - g(a)}g^{\prime}(c) = 0[/math][/center][br]Reorganizing terms gives us[br][br][center][math]\dfrac{f^{\prime}(c)}{g^{\prime}(c)} = \dfrac{f(b) - f(a)}{g(b) - g(a)}[/math][/center][br][br]This completes the proof.