I must apply the Rollle's theorem to $f(x)=x(x^2-4)$ in $[0,1]$ and $[-1,1]$.

But if I want to pull this theorem I must just verify that the function is continuous in the closed interval and differentiable in the open interval and in the extreme values the function has the same value, however in this particular function we have that is continuous and differentiable but $f(0)= 0$,$f(1)=-3$,$f(-1)=3$ then how can I apply the Rolle's theorem.

Thanks for your time and help.

  • $\begingroup$ Maybe use Rolle's theorem to to show there isn't a point in those intervals where the slope is zero $\endgroup$ – Triatticus Apr 19 '16 at 19:16
  • $\begingroup$ As a counterexample? $\endgroup$ – Rachel Apr 19 '16 at 19:17
  • $\begingroup$ I'm simply guessing as you are right that rolled theorem seems not to apply, I would have used it to show the function doesn't have a stationary point inside the intervals in question, which is true because it is a cubic with extrema outside the interval $\endgroup$ – Triatticus Apr 19 '16 at 19:19

You can only apply the contrapositive of Rolle's Theorem here.

First note that $f(x) = x(x^2 - 4)$ has stationary points at $x = + \frac{2\sqrt{3}}{3}$ and $x = - \frac{2\sqrt{3}}{3}$. But neither of these are in the intervals $[0, 1]$ or $[-1, 1]$.

So, by the contrapositive of Rolle's Theorem, there is no repeated value of $f(x)$ in the interval $[0, 1]$ and $[-1 ,1]$.


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