zero position

Mar 2010
91
16
usa
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 \(\displaystyle c-a > b-a\)
 
Mar 2010
41
18
unfortunalty, im stuck in hicks state. IN.
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 \(\displaystyle c-a > b-a\)
c-a>b-a is just c>b, but b is the right endpoint
So this seems wrong.
maybe c-a>b-c?