91Ó°ÊÓ

Critical points are essential in calculus when determining where a function might achieve its highest or lowest values. They occur at values of \(x\) where the derivative, or slope of the tangent line to the function, equals zero. These points are "critical" because a change from increasing to decreasing function values, or vice-versa, might happen at these points. To find these critical points, first, compute the derivative of the function. For the function \(f(x) = x^4 - 2x^2 + 3\), the derivative is \(f'(x) = 4x^3 - 4x\). Set this derivative equal to zero:

• \(4x(x^2 - 1) = 0\)

Solving this equation, you get\(x = 0\), \(x = 1\), and \(x = -1\) as critical points. These points, along with the endpoints of the interval, are potential candidates for absolute extrema.

The first derivative \(f'(x)\) of a function describes the rate of change of the function's values and gives the slope of the tangent line at any point \(x\). When finding maximum and minimum values of a function on an interval, the first derivative is crucial. In our exercise, after finding \(f'(x) = 4x^3 - 4x\), the solution involved setting it to zero to find critical points, which can indicate potential maximum or minimum points:

• \(f'(x) = 4x(x^2 - 1) = 0\)

This leads to the solutions \(x = 0\), \(x = 1\), and \(x = -1\).
The critical points obtained from the first derivative test will be evaluated, alongside the endpoints of the interval, to determine the absolute maximum and minimum values.

The final step in finding absolute maximum and minimum values involves evaluating the original function not just at the critical points, but also at the interval's endpoints. This ensures that the highest and lowest values on the entire interval are identified. For the given function \(f(x) = x^4 - 2x^2 + 3\) on the interval \([-2, 3]\), we evaluate:

• \(f(-2) = 16 - 8 + 3 = 11\)
• \(f(3) = 81 - 18 + 3 = 66\)

Together with critical point evaluations:

• \(f(-1) = 2\)
• \(f(0) = 3\)
• \(f(1) = 2\)

Comparing all these values, the absolute maximum is \(66\) at \(x = 3\) and the absolute minimum is \(2\), occurring at \(x = -1\) and \(x = 1\). By checking endpoints and critical points, we've ensured that no potential extremum has been overlooked.