# Mean Value Theorem

• Nov 18th 2008, 06:26 PM
Fosite
Mean Value Theorem
Suppose that $g$ is a function defined and continuous on [-2, 2] and that $g"$ exists on the
open interval (-2, 2) . If $g(-2) = 5$ , $g(2) = 1$ and $g"(x)>= 4$ for all $x$ in (-2, 2) , how
large can $g(1)$ possibly be?

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

$g(1)-g(-2)=g"(c)(1+2)$
$g(1)=g(-2)+3g"(c)$
$3g"(c)>=12$
$g(1)=5+3g"(c)>=17$

so is $g(1)$ infinity ?

and then..

$g(1)-g(2)=g"(c)(1-2)$
$g(1)=g(2)-g"(c)$
$-g"(c)=<4$
$g(1)=1-g"(c)=<-3$

I got confused here...

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

and then I tried another way...

$g(2)-g(1)=g'(c)(2-1)$
$g(-2)-g(1)=g'(c)(-2-1)$
$1-g(1)=g'(c)$
$5-g(1)=-3g'(c)$
$g(1)=2 g'(c)=-1$

Think geometrically! The condition $g''(x)\geqslant4$ 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 $g(x) = 2x^2+ax+b$. 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.