91Ó°ÊÓ

Begin by looking at the given function and simplifying it as much as possible to make the task a bit easier. The given function is \(f(x) = \left|\frac{x^2-2}{x^2-4}\right|\). Since the absolute value is always non-negative, we can simplify the expression inside the absolute value signs: \(f(x) = \left|\frac{x^2-2}{(x+2)(x-2)}\right|\).

Now, we'll find the first derivative of the given function, which will help us identify any critical points. Applying quotient rule yields \(f'(x) = -\frac{2x^3 - 12x}{(x^2 - 4)^2}\).

To find the critical points, we need to find where \(f'(x) = 0\) or it is undefined. Since the denominator is never equal to 0 within the absolute value, it only occurs when the numerator is 0. In this case, we have \(2x^3 - 12x = 0\). Factor out 2x: \(2x(x^2 - 6) = 0\). This leads to three critical points: \(x = 0\), \(x = \sqrt{6}\), and \(x = -\sqrt{6}\). Now, let's analyze the behavior around the critical points to identify whether they are a maxima or minima. Since the function is within an absolute value, there can't be any local minima as the function can't decrease from any positive value. So, let's try to find out if any of these critical points are local maxima. Notice that when \(x<0\), the absolute value in the function \(f(x)\) results in a change of sign in the expression \(\frac{x^2-2}{(x+2)(x-2)}\). Therefore, \(-\sqrt{6} < 0\) will result in a decrease in the function values. Meanwhile, \(0\) and \(\sqrt{6}\) will result in increases in the function values, so these points can't be maxima. Therefore, option (B) is the correct answer, as there are no points of local maxima.