
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..
$\displaystyle g(1)g(2)=g"(c)(12)$
$\displaystyle g(1)=g(2)g"(c)$
$\displaystyle g"(c)=<4$
$\displaystyle g(1)=1g"(c)=<3$
I got confused here...
will $\displaystyle f(b)f(a)=f"(c)(ba)$ ???
and then I tried another way...
$\displaystyle g(2)g(1)=g'(c)(21)$
$\displaystyle g(2)g(1)=g'(c)(21)$
$\displaystyle 1g(1)=g'(c)$
$\displaystyle 5g(1)=3g'(c)$
$\displaystyle g(1)=2 g'(c)=1$
Am I wrong again???(Headbang)

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.