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  • https://math.libretexts.org/Bookshelves/Analysis/Introduction_to_Mathematical_Analysis_I_(Lafferriere_Lafferriere_and_Nguyen)/04%3A_Differentiation/4.06%3A_Section_6-
    \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.

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