1. ## Mean Value Theorem

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???

2. Think geometrically! The condition $\displaystyle g''(x)\geqslant4$ tells you that the function g is convex. The maximum value of g(1) will occur when g''(x) is equal to 4 throughout the interval. If g''(x) is ever greater than 4 (with the endpoints at (-2,5) and (2,1) remaining fixed) then the curve will become more convex, forcing g(1) to be smaller.

So let g''(x) = 4. Integrate twice to get $\displaystyle g(x) = 2x^2+ax+b$. You can find the values of the constants a and b by using the facts that g(-2)=5 and g(2)=1, and then you will know what g(1) is.