Let f(x) be a differentiable function and $\mathrm{f}\left(\mathrm{a}\right)=\mathrm{f}\left(\mathrm{b}\right)=0$ $\left(\mathrm{\alpha }<\mathrm{\beta }\right)$, then in the interval (a, b),

Consider $\mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right).$ 

Then g(a) = g(b).

using Rolle's theorem, there exists at least one $\mathrm{c}\in \left(\mathrm{a},\mathrm{b}\right)$ for which

$\begin{array}{l} \mathrm{g}\text{'}\left(\mathrm{c}\right)=0\\ \mathrm{or} {\mathrm{e}}^{\mathrm{c}}\mathrm{f}\left(\mathrm{c}\right)+{\mathrm{e}}^{\mathrm{c}}\mathrm{f}\text{'}\left(\mathrm{c}\right)=0\\ \mathrm{or} \mathrm{f}\left(\mathrm{c}\right)+\mathrm{f}\text{'}\left(\mathrm{c}\right)=0\end{array}$

Thus, f(x) + f' (x) = 0 has at least one root in (a, b).

Consider $\mathrm{g}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)$

then g(a) = g(b).

Using Rolle's theorem, there exists at least one $\mathrm{c}\in \left(\mathrm{a},\mathrm{b}\right)$ for which

$\begin{array}{l} \mathrm{g}\text{'}\left(\mathrm{c}\right)=0\\ \mathrm{or} -{\mathrm{e}}^{-\mathrm{c}}\mathrm{f}\left(\mathrm{c}\right)+{\mathrm{e}}^{\mathrm{c}}\mathrm{f}\text{'}\left(\mathrm{c}\right)=0\\ \mathrm{or} -\mathrm{f}\left(\mathrm{c}\right)+\mathrm{f}\text{'}\left(\mathrm{c}\right)=0\end{array}$

$\begin{array}{l}\mathrm{Thus}, \mathrm{f}\text{'}\left(\mathrm{x}\right)-\mathrm{f}\left(\mathrm{x}\right)=0 \mathrm{has} \mathrm{at} \mathrm{least} \mathrm{one} \mathrm{root} \mathrm{in} \left(\mathrm{a},\mathrm{b}\right).\\ \mathrm{Now}, \mathrm{consider} \mathrm{g}\left(\mathrm{x}\right)={\left(\mathrm{f}\left(\mathrm{x}\right)\right)}^2.\\ \mathrm{Then} \mathrm{g}\left(\mathrm{a}\right)=\mathrm{g}\left(\mathrm{b}\right).\end{array}$

Using Rolle's theorem, there exists at least one $\mathrm{c}\in \left(\mathrm{a},\mathrm{b}\right)$ for which

$\begin{array}{l} \mathrm{g}\text{'}\left(\mathrm{c}\right)=0\\ \mathrm{or} 2\mathrm{f}\text{'}\left(\mathrm{c}\right)\mathrm{f}\left(\mathrm{c}\right)=0\\ \mathrm{or} \mathrm{f}\text{'}\left(\mathrm{c}\right)\mathrm{f}\left(\mathrm{c}\right)=0\end{array}$

Thus, $\mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{f}\left(\mathrm{x}\right)=0$ has at least one root in (a, b).