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Critical points are particular values in the domain of a function where the function's derivative is zero or does not exist. These points are important because they often correspond to local maximum or minimum values of the function.
Finding critical points involves:

• Taking the derivative of the function.
• Setting the derivative equal to zero.
• Solving for the variable.

However, sometimes the derivative is never zero within the given interval. In our exercises with the functions \( f(x) = \frac{x+1}{x-1} \) and \( g(t) = \frac{t}{t-1} \), the derivatives are \( f'(x) = \frac{-2}{(x-1)^2} \) and \( g'(t) = \frac{-1}{(t-1)^2} \), respectively. Since these ratios are never zero, there are no critical points within the interval \([2,4]\).
This means we only need to consider values at the endpoints of the interval, simplifying the search for absolute extrema.

The absolute maximum of a function on a given interval is the highest point within that interval. It is the largest value of the function, meaning no other point in the interval has a higher value.
Here's how to determine the absolute maximum:

• Evaluate the function at its critical points (if any).
• Evaluate the function at the endpoints of the interval.
• Compare these values and identify the greatest one.

When we evaluated \( f(x) = \frac{x+1}{x-1} \) on \([2, 4]\), we found:
\( f(2) = 3 \)
\( f(4) = \frac{5}{3} \)
Thus, the absolute maximum is \( 3 \) at \( x = 2 \). Similarly, for \( g(t) = \frac{t}{t-1} \) evaluated on \( [2, 4] \), the absolute maximum is \( 2 \) at \( t = 2 \).
This process of checking endpoints alongside critical points ensures you capture the highest function output possible on a closed interval.

The absolute minimum of a function over a given interval is the lowest point within that interval. It is the smallest value of the function, meaning no other point in the interval has a lesser value.
Here's how to identify the absolute minimum:

• Evaluate the function at its critical points (if any).
• Evaluate the function at the endpoints of the interval.
• Compare these values and identify the smallest one.

From our calculations for \( f(x) = \frac{x+1}{x-1} \) on \([2, 4]\), we determined that:
\( f(2) = 3 \)
\( f(4) = \frac{5}{3} \)
The absolute minimum is \( \frac{5}{3} \) at \( x = 4 \). Similarly, for \( g(t) = \frac{t}{t-1} \), the evaluated values gave us an absolute minimum of \( \frac{4}{3} \) at \( t = 4 \).
Identifying absolute minima involves comparing function values at specific key points, ensuring the smallest possible output for the function is found across the interval.