If the functions f(x) and g(x) are continuous on [a, b] and differentiable on (a, b), then in the interval (a, b), the equation

$\left|\begin{array}{ll} f^{\prime}(x) & f(a) \\ g^{\prime}(x) & g(a) \end{array}\right|=\frac{1}{a-b}\left|\begin{array}{ll} f(a) & f(b) \\ g(a) & g(b) \end{array}\right|$

has at least one root

Consider the function $\phi(x)$ given by

$\phi(x)=\left|\begin{array}{ll} f(a) & f(x) \\ g(a) & g(x) \end{array}\right|$

Since $f(x)$ and $g(x)$ are continuous on $[a, b]$ and differentiable on $(a, b)$. Therefore, $\phi(x)$ is continuous on $[a, b]$ and differentiable on $(a, b)$. Consequently, by Lagrange's mean value theorem there exists at least one $c \in(a, b)$ such that

$\phi^{\prime}(c)=\frac{\phi(b)-\phi(a)}{b-a}$

$\Rightarrow\left|\begin{array}{ll} f(a) & f'(c) \\ g(a) & g'(c) \end{array}\right|=\frac{1}{b-a}\left|\begin{array}{ll} f(a) & f(b) \\ g(a) & g(b) \end{array}\right|$

$\Rightarrow\left|\begin{array}{ll} f^{\prime}(c) & f(a) \\ g^{\prime}(c) & g(a) \end{array}\right|=\frac{1}{a-b}\left|\begin{array}{ll} f(a) & f(b) \\ g(a) & g(b) \end{array}\right|$

Hence, the equation

$\left|\begin{array}{ll} f^{\prime}(x) & f(a) \\ g^{\prime}(x) & g(a) \end{array}\right|=\frac{1}{a-b}\left|\begin{array}{ll} f(a) & f(b) \\ g(a) & g(b) \end{array}\right|$

has at least one root in (a, b).