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The given function is \(f(t)=\frac{t^{2}+1}{t^{2}-1}\). Notice that the function is undefined at \(t = \pm 1\), so we need to keep that in mind when solving for extrema. The domain of this function is \(D = (-\infty, -1) \cup (-1, 1) \cup (1, \infty)\). Since we are only considering the interval \(-2 \leq t \leq 2\), we will focus on the subintervals \([-2, -1) \cup (-1, 1) \cup (1, 2]\).

To find critical points, we must take the first derivative of the function and solve for points where the derivative is either equal to zero or undefined. First, we find the derivative of the function: \(f'(t) = \frac{d}{dt}(\frac{t^2+1}{t^2-1}) = \frac{2t(t^2-1) - 2t(t^2+1)}{(t^2-1)^2}\) Next, we look for points where the derivative is zero or undefined: 1. Solving for values of \(t\) where \(f'(t) = 0\) To find out when the derivative is zero, we just need to analyze the numerator of the derivative (as the denominator will not impact the zeros): \(2t(t^2-1) - 2t(t^2+1) = 0\) Solving this equation gives no real solutions within our interval. Therefore, there are no points where the derivative is zero. 2. Solving for values of \(t\) where \(f'(t)\) is undefined We know that the derivative is undefined at \(t = \pm 1\), which are also the points of discontinuity in the function. Thus, we have two critical points, \(t = -1\) and \(t = 1\).

To determine whether the critical points correspond to relative extrema, we analyze the behavior of the first derivative around the critical points for changes in sign (which indicate a local maximum or minimum). 1. Interval \((-2, -1)\): The sign of the derivative is positive, indicating an increasing function on this interval. 2. Interval \((-1, 1)\): The derivative is negative, indicating a decreasing function on this interval. 3. Interval \((1, 2)\): The derivative is positive, indicating an increasing function on this interval. We can see that there is a change in the behavior of the function at the critical points: the function starts increasing in the first interval, then decreases in the second interval and finally increases in the third interval. This indicates that there is a relative maximum at \(t = -1\) and a relative minimum at \(t = 1\). However, since these points are not in the domain of the function, they cannot be considered relative extrema. Next, we evaluate the function at the endpoints, -2 and 2, to check for the existence of absolute extrema. \(f(-2) = \frac{(-2)^2+1}{(-2)^2-1} = \frac{5}{3}\) \(f(2) = \frac{(2)^2+1}{(2)^2-1} = \frac{5}{3}\) Since the function has the same value at -2 and 2 and does not attain any extreme points within the domain, there are no relative extrema in the given interval. Therefore, the function does not have any relative or absolute extrema in the given interval \(-2 \leq t \leq 2\).