\end{array}\] Since \(f^{\prime}\left(c_{1}\right) \leq f^{\prime}\left(c_{2}\right)\), we have \[t f\left(x_{t}\right)-t f\left(x_{1}\right)=f^{\prime}\left(c_{1}\right) t(1-t)\left(x_{2}-x_{1}\right...\end{array}\] Since \(f^{\prime}\left(c_{1}\right) \leq f^{\prime}\left(c_{2}\right)\), we have \[t f\left(x_{t}\right)-t f\left(x_{1}\right)=f^{\prime}\left(c_{1}\right) t(1-t)\left(x_{2}-x_{1}\right) \leq f^{\prime}\left(c_{2}\right) t(1-t)\left(x_{2}-x_{1}\right)=(1-t) f\left(x_{2}\right)-(1-t) f\left(x_{t}\right) .\] Rearranging terms, we get \[f\left(x_{t}\right) \leq t f\left(x_{1}\right)+(1-t) f\left(x_{2}\right) .\] Therefore, \(f\) is convex.