# Mean Value Theorem

• Nov 18th 2008, 06:26 PM
Fosite
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\$