 Solution

The derivative is $f^{\prime }(x)=2x\cos (x^2-2)$. Thus a stationary point occurs either at $x=0$ or where $x^2-2=\pi /2+k\pi$. Since $-2\le x^2-2<2$ on the specified open interval, the only allowed cosine zeros are $-\pi /2$ and $\pi /2$. The five stationary abscissae areTheir ordinates are $-\sin 2$, $-1$, and $1$, respectively. For completeness, $f^{\prime \prime }=2\cos (x^2-2)-4x^2\sin (x^2-2)$ shows that the inner pair are local minima and the outer pair are local maxima; $x=0$ is also a local maximum because $2\cos 2<0$.