Let k=(f(b)-f(a))/(b-a)

Because this function is nonlinear, x=c∈(a, b) exists.

f(c)≠f(a)+k(c-a)

Suppose f(c) > female (female)+male (female)

Using lagrange mean value theorem in (a, c)

get f '(η)=(f(c)-f(a))/(c-a)>； k=(f(b)-f(a))/(b-a)

| f '(η)| & gt； |(f(b)-f(a))/(b-a)|

f(c)& lt； F(a)+k(c-a) also holds.

Eleven questions will not. After all, I'm not a math major.