Stationary Points

First we find the derivative $f'\left( x \right)$ :

$f'\left( x \right)={ x }^{ 2 }+2ax-2.$

Solving the equation $f'\left( x \right)=0$ yields the x-coordinates of the stationary points of $f$ :

$\Rightarrow { x }^{ 2 }+2ax-2=0\\ \Rightarrow { (x+a) }^{ 2 }-{ a }^{ 2 }-2=0\\ \Rightarrow { (x+a) }^{ 2 }={ a }^{ 2 }+2\\ \Rightarrow x=-a\pm \sqrt { { a }^{ 2 }+2 }.$

Hence, $x=-a+\sqrt { { a }^{ 2 }+2 }$ and $x=-a-\sqrt { { a }^{ 2 }+2 }$ .

Consider the point $x=-a+\sqrt { { a }^{ 2 }+2 }$ . This expression will always be a positive value, which can be clearly seen from the graphs of $a$ and $\sqrt { { a }^{ 2 }+2 }$ . Let $y=\sqrt { { a }^{ 2 }+2 }$ .

$\Rightarrow \frac { { y }^{ 2 } }{ 2 } -\frac { { a }^{ 2 } }{ 2 } =1$ which is the top half of a hyperbola with asymptote $y=a$ . Hence, $\sqrt { { a }^{ 2 }+2 }>a$ and $-a+\sqrt { { a }^{ 2 }+2 }>0$ .

Now consider the point $x=-a-\sqrt { { a }^{ 2 }+2 }$ . We have already shown that $\sqrt { { a }^{ 2 }+2 }>a$ for all $a$ . As $\sqrt { { a }^{ 2 }+2 }$ is an even function, this implies that $\sqrt { { a }^{ 2 }+2 }>-a$ .

$\Rightarrow -a-\sqrt { { a }^{ 2 }+2 } <0$

And hence the second stationary point will always occur at a negative value of $x$ . Since we have the first stationary point always positive for $x$ and the second always negative, they will never be in the same quadrant of the cartesian plane, and the probability is simply $\boxed { 0 }$ .