Ask your own question, for FREE!
Mathematics 6 Online
OpenStudy (darkprince14):

help: please correct my proof

OpenStudy (darkprince14):

OpenStudy (thomas5267):

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.

OpenStudy (thomas5267):

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!
Latest Questions
Mari103: How to pop out like a Jacc In the box
7 hours ago 0 Replies 0 Medals
Breathless: Spooky witch but cute
14 hours ago 3 Replies 0 Medals
Arriyanalol: help
13 hours ago 10 Replies 2 Medals
Arriyanalol: @tinydinoUwU stop trying to find a argument u blad lil boy
1 day ago 5 Replies 4 Medals
Jaded012023: Please tell me what you all think of this song
16 hours ago 6 Replies 1 Medal
Arriyanalol: bro how
16 hours ago 2 Replies 3 Medals
Arriyanalol: cant wait for the new bluey movie in 2027
1 day ago 12 Replies 2 Medals
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!