Ask
your own question, for FREE!
Mathematics
6 Online
help: please correct my proof
Still Need Help?
Join the QuestionCove community and study together with friends!
I don't know but I think you shouldn't assume that f is differentiable. I suspect this can be solved using intermediate value theorem only.
Let \(g(z)=cf(x)+kf(y)-(c+k)f(z)\). \(g(z)\) is continuous because it is a combination of the continuous function \(f(x)\). Suppose \(f(x)<f(y)\). Then \(g(y)<0\) and \(g(x)>0\). By Intermediate Value Theorem, there exist a \(\zeta\) such that \(x<\zeta<y\) and \(g(\zeta)=0\). \(g(\zeta)=0\implies f(\zeta)=\dfrac{cf(x)+kf(y)}{c+k}\). The case of \(f(y)<f(x)\) and \(f(x)=f(y)\) is left as an exercise for the reader. The case of \(f(x)=f(y)\) is slightly harder, and that will be my belated gift for you on April Fool's day.
Can't find your answer?
Make a FREE account and ask your own questions, OR help others and earn volunteer hours!
Join our real-time social learning platform and learn together with your friends!
Join our real-time social learning platform and learn together with your friends!
Latest Questions
Arriyanalol:
@tinydinoUwU stop trying to find a argument u blad lil boy
TinydinoUwU:
**(Verse 1)** Yo, trapped in a box, Iu2019m feelin' so confined, Lifeu2019s a game of chess, but Iu2019m stuck in rewind, Every dayu2019s a struggle, man, I
Arriyanalol:
hey umm so i need help with my lanauage art ixl anybody wanna help big mama
Nina001:
ho where do i go to buy Subscirption for a moving pfp because on my screen im on
1 day ago
5 Replies
4 Medals
16 hours ago
13 Replies
5 Medals
4 days ago
2 Replies
2 Medals
5 days ago
4 Replies
2 Medals