Find out the interval where Rolle's Theorem is applicable Find out the interval for which the Rolle's theorem is valid for the function    $f(x)=2x^3+x^2-4x+2$   

My attempt : Supposing the interval is $[a,b]$, $f(a)=f(b)$ gives the equation $2(a^2+ab+b^2)+a+b-4=0$. Putting b=-a, we can get the interval to be $[-\sqrt 2, \sqrt 2]$. 

Is there any way to do it directly?
 A: Your idea is fine.
The procedure could be shortened by asking directly for a $b>0$ with $f(-b)=f(b)$. This means that the odd part $2x^3-4x$ of $f$ should vanish at $b$. This immediately leads to $b=\sqrt{2}$.
A: This is not true. Let $ b=t> a $; the general solution can be found by solving the equation $f (t)-f (a)=0$ for the variable $ t $. After simplifying,  we get three roots $$ t=a, -\frac{1}{2} a-\frac {1}{4}\pm\frac {1}{4}\sqrt {  -12a^2-4a+33  }.$$ of course the first solution $ t=a $ rejected because we have interval. It remains to check the validity of the 2nd and 3rd roots by solving the quadratic inequality  $-12a^2-4a+33\ge 0$ simple calculation yields the real solution is $ a\in [-\frac {11}{6},\frac {3}{2}]$. Therefore,  for the 2nd solution $\cdots +1/4\sqrt {. . .} $ to be $> a $ (because we seek interval) we have to choose $a\in [\frac {-11}{6},\frac {3}{2})$. Similarly, for the 3rd solution $\cdots -1/4\sqrt {. . . }$ to be $> a$ we must choose $ a\in [-\frac {11}{6}, -1) $.
