- The derivative \(f'(x) = 2x - 2\) identifies the critical point at \(x = 1\) with a function value of \(f(1) = -1\).
- On the interval \([-1, 2]\), comparing values at boundaries and the critical point gives an absolute maximum of \(3\) at \(x = -1\) and an absolute minimum of \(-1\) at \(x = 1\).
- On the interval \((1, 3]\), since the function is strictly increasing, the absolute maximum is \(3\) at \(x = 3\), and there is no absolute minimum.
- Final results:
- \([-1, 2]\): Max \(3\) at \(x = -1\), Min \(-1\) at \(x = 1\).
- \((1, 3]\): Max \(3\) at \(x = 3\), Min: none.
Explanation
- Analysis of the Function and Critical Points
The function \(f(x) = x^2 - 2x\) is a downward-opening parabola? No, it's an upward-opening parabola with a vertex at \(x = 1\). We calculate the derivative, \(f'(x) = 2x - 2\), to find the critical points. Setting \(f'(x) = 0\) gives \(2x - 2 = 0\), so the only critical point is \(x = 1\). The function value at the vertex is \(f(1) = 1^2 - 2(1) = -1\). For any interval, the absolute extrema can occur either at the critical points within the interval or at the boundaries of the interval.
- Absolute Extrema on \([-1, 2]\)
For the interval \([-1, 2]\), we evaluate the function at the boundaries \(x = -1\) and \(x = 2\), as well as the critical point \(x = 1\) since it lies within the interval.
- At \(x = -1\): \(f(-1) = (-1)^2 - 2(-1) = 1 + 2 = 3\).
- At \(x = 1\): \(f(1) = (1)^2 - 2(1) = -1\).
- At \(x = 2\): \(f(2) = (2)^2 - 2(2) = 4 - 4 = 0\).
Comparing these values \(\{3, -1, 0\}\), the largest value is \(3\) and the smallest is \(-1\). Thus, the absolute maximum is \(3\) at \(x = -1\) and the absolute minimum is \(-1\) at \(x = 1\).
- Absolute Extrema on \((1, 3]\)
For the interval \((1, 3]\), the endpoint \(x = 1\) is excluded. The derivative \(f'(x) = 2x - 2\) is positive for all \(x > 1\), meaning the function is strictly increasing on this interval.
- As \(x\) approaches \(1\) from the right, the function value \(f(x)\) approaches \(f(1) = -1\), but never reaches it because \(1\) is excluded.
- At the endpoint \(x = 3\): \(f(3) = 3^2 - 2(3) = 9 - 6 = 3\).
Because the function is strictly increasing, it attains its absolute maximum at the largest value in the interval, which is \(x = 3\), and it has no absolute minimum because the value \(-1\) is never attained.
- Summary of Results
In summary:
- On the interval \([-1, 2]\), the absolute maximum is \(3\) at \(x = -1\) and the absolute minimum is \(-1\) at \(x = 1\).
- On the interval \((1, 3]\), the absolute maximum is \(3\) at \(x = 3\), and there is no absolute minimum.
Examples
In manufacturing, if a machine's power output during a cycle is modeled by a function like \(f(x) = x^2 - 2x\), engineers use this to find the peak performance (absolute maximum) and the lowest power state (absolute minimum) during specific time intervals. This ensures the machine operates within safe limits and optimizes efficiency during production runs. Understanding these extrema helps in calibrating equipment to prevent damage or failure when the function reaches its boundaries.