Suppose that $\displaystyle g$ is a function defined and continuous on [-2, 2] and that $\displaystyle g"$ exists on the

open interval (-2, 2) . If $\displaystyle g(-2) = 5$ , $\displaystyle g(2) = 1$ and $\displaystyle g"(x)>= 4$ for all $\displaystyle x$ in (-2, 2) , how

large can $\displaystyle g(1)$ possibly be?

I did the follow step, but something goes wrong..

$\displaystyle g(1)-g(-2)=g"(c)(1+2)$

$\displaystyle g(1)=g(-2)+3g"(c)$

$\displaystyle 3g"(c)>=12$

$\displaystyle g(1)=5+3g"(c)>=17$

so is $\displaystyle g(1)$ infinity ?

and then..

$\displaystyle g(1)-g(2)=g"(c)(1-2)$

$\displaystyle g(1)=g(2)-g"(c)$

$\displaystyle -g"(c)=<4$

$\displaystyle g(1)=1-g"(c)=<-3$

I got confused here...

will $\displaystyle f(b)-f(a)=f"(c)(b-a)$ ???

and then I tried another way...

$\displaystyle g(2)-g(1)=g'(c)(2-1)$

$\displaystyle g(-2)-g(1)=g'(c)(-2-1)$

$\displaystyle 1-g(1)=g'(c)$

$\displaystyle 5-g(1)=-3g'(c)$

$\displaystyle g(1)=2 g'(c)=-1$

Am I wrong again???