<s>[INST] An application of the Mean Value Theorem I'm recalling this question from memory, so I may be messing it up a bit.
Let $a/3+b/2+c=0$.  Show that $ax^2+bx+c=0$ has at least one root in $[0,1]$ using the Mean Value Theorem.
Let $f(x)=ax^2+bc+c$.  Then $f(0)=c$ and $f(1)=a+b+c$.  Also $f'(x)=2ax+b$.  So there exists $f(\xi)=[f(1)-f(0)]/1=a+b-c$.  Then $a+b-c=2a\xi+b \Rightarrow (a-c)/2a=\xi$.
I'm not sure if this is right or where to go from here.
 [/INST] First, if $a =0$, then we have $bx + c = 0 \implies x = - \frac{c}{b} = \frac{b/2}{b} = \frac{1}{2}$.
Now, suppose $a \neq 0$.
Note that $c = - \frac{a}{3} - \frac{b}{2}$, so you want to prove that the function $f(x) = ax^2 + bx - \frac{a}{3} - \frac{b}{2}$ has a root in $[0,1]$.  We have $f(0) = - \frac{a}{3} - \frac{b}{2}$ and $f(1) =  \frac{2}{3} a + \frac{1}{2} b$.  Note that 
$$f(0)\cdot f(1) = - \frac{2}{9}a^2 - \frac{1}{4}b^2 < 0$$ as $a \neq 0$.  Thus, $f(0)$ and $f(1)$ have different signs, and by the Intermediate Value Theorem, there is a root in $[0,1]$.
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