\end{array}\] Since f′(c1)≤f′(c2), 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′(c1)≤f′(c2), we have tf(xt)−tf(x1)=f′(c1)t(1−t)(x2−x1)≤f′(c2)t(1−t)(x2−x1)=(1−t)f(x2)−(1−t)f(xt). Rearranging terms, we get f(xt)≤tf(x1)+(1−t)f(x2). Therefore, f is convex.
