Question

If Rolle's theorem holds for the function $f\left(x\right)=x^3-4x+1,x\in [-2,2]$ at the point $x=c$, then the value of $3c^2$ is:

Answer 1

First we make sure that we can apply Rolle's theorem. As we deal with a polynomial, this function is continuous and differentiable in the given interval. Computing the values at the endpoints of the interval:
$f(-2)=(-2{)}^3-4\cdot (-2)+1=1,$
$f\left(2\right)=2^3-4\cdot 2+1=1$
Thus, the function has equal values at the endpoints. Hence, all three conditions of Rolle's theorem hold.
Now, to find the values of $c$ we apply Rolle's theorem.
$f^{\prime }\left(x\right)=(x^3-4x+1{)}^{\prime }=3x^2-4.$
$f^{\prime }\left(c\right)=0,\Rightarrow 3c^2-4=0,\Rightarrow c^2=\frac{4}{3},\Rightarrow c_{1,2}=\pm \frac{2}{\sqrt{3}}=\pm \frac{2\sqrt{3}}{3}\approx \pm 1.15$
So we obtained two values $c_1=-\frac{2\sqrt{3}}{3}$ and $c_2=\frac{2\sqrt{3}}{3}$ where the derivative of the function is zero. Both of them belong to the open interval $(-2,2).$
$\therefore$ We have $3c^2=4$