Question

The point(s) $c$ satisfying the conditions of the LMVT for the function $f\left(x\right)=x^3-x$ in the interval $[-2,1]$ is:

• A. $\pm \sqrt{3}$
  
• B. $\pm 1$
• C. $-1$
• D. $\pm \frac{1}{\sqrt{3}}$
  

Answer 1

The correct option is C $-1$

We have here a cubic function which is continuous on the closed interval $[-2,1]$ and differentiable on the open interval $(-2,1)$. So, using LMVT we have:
$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$
$=\frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{(1^3-1)-\left(\right(-2{)}^3-(-2))}{1-(-2)}$
$=\frac{6}{3}=2$

Now, $f^{\prime }\left(x\right)=(x^3-x{)}^{{}^{\prime }}=3x^2-1$
So, we have:
$3c^2-1=2\Rightarrow c^2=1\Rightarrow c=\pm 1$
$\therefore$The points $c$ satisfying the conditions of the LMVT is $-1$