E elim Mar 2010 91 16 usa May 15, 2010 #1 Lef \(\displaystyle f'' > 0\) on \(\displaystyle [a,b]\) with \(\displaystyle f(a) < 0 = f(c) < f(b), \quad \int_a^b f(x)dx = 0\) Show that \(\displaystyle c-a > b-a\)

Lef \(\displaystyle f'' > 0\) on \(\displaystyle [a,b]\) with \(\displaystyle f(a) < 0 = f(c) < f(b), \quad \int_a^b f(x)dx = 0\) Show that \(\displaystyle c-a > b-a\)

autumn Mar 2010 41 18 unfortunalty, im stuck in hicks state. IN. May 15, 2010 #3 elim said: Lef \(\displaystyle f'' > 0\) on \(\displaystyle [a,b]\) with \(\displaystyle f(a) < 0 = f(c) < f(b), \quad \int_a^b f(x)dx = 0\) Show that \(\displaystyle c-a > b-a\) Click to expand... c-a>b-a is just c>b, but b is the right endpoint So this seems wrong. maybe c-a>b-c?

elim said: Lef \(\displaystyle f'' > 0\) on \(\displaystyle [a,b]\) with \(\displaystyle f(a) < 0 = f(c) < f(b), \quad \int_a^b f(x)dx = 0\) Show that \(\displaystyle c-a > b-a\) Click to expand... c-a>b-a is just c>b, but b is the right endpoint So this seems wrong. maybe c-a>b-c?

E elim Mar 2010 91 16 usa May 15, 2010 #4 autumn said: c-a>b-a is just c>b, but b is the right endpoint So this seems wrong. maybe c-a>b-c? Click to expand... You are right, sorry about the typo

autumn said: c-a>b-a is just c>b, but b is the right endpoint So this seems wrong. maybe c-a>b-c? Click to expand... You are right, sorry about the typo