Dada la funcion f(x)=x^3-4x defina sobre el intervalo[-2,2] hallar el valor c ∈(-2,2) que satisface f '(c)=0

Oosj Estrada!

$$\begin{align}&\ y=x^3-4x\\ &y'=3x^2-4\\ &y'=0 \Rightarrow3x^2-4=0 \Rightarrow x^2=\frac{4}{3}\\ & x=\pm \sqrt{ \frac{4}{3}}= \pm \frac{2}{\sqrt{3}}= \pm1,154....\\ &\\ &c=\pm1,154....\end{align}$$