91Ó°ÊÓ

In calculus, critical points are essential for identifying where a function might achieve its local maxima or minima. These are points where the derivative of a function either equals zero or does not exist. When working with a function like \(f(x)\), finding these points is an important step in optimization problems. Here is how you find them:

• First, compute the derivative of the function. For a function \(f(x)\), you're looking for \(f'(x)\).
• Next, solve \(f'(x) = 0\) to find values of \(x\) where the slope of the tangent is zero.
• Consider also where \(f'(x)\) might not exist, which could also indicate potential critical points.

Critical points serve as the foundation for identifying the behavior of the function around certain values of \(x\). In our exercise, the critical point was found to be \(x = 1\). This is inside the given interval and thus important for subsequent analysis. Remember, only the critical points that fall within the problem's interval are typically useful for further steps in optimization.

The derivative is a core concept in calculus, often called the rate of change of a function. When solving problems involving optimization, the derivative helps us understand how a function behaves. To find a derivative like in this exercise:

• Use rules such as the quotient rule if dealing with a fraction as \(f(x) = \frac{x}{x^2 - x + 1}\).
• The quotient rule states: If \(f(x) = \frac{u(x)}{v(x)}\), then \(f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}\).
• Apply this rule to get \[f'(x) = \frac{(1)(x^2 - x + 1) - x(2x - 1)}{(x^2-x+1)^2}\].

Solving for where this derivative equals zero will help locate critical points. Simplifying derivatives often involves algebraic manipulation, which can be tricky, but is crucial for exactness in subsequent steps.

Identifying the absolute maximum and minimum values of a function in a defined interval is the core goal of an optimization problem. Here’s how it’s typically done:

• Find the critical points of the function within the interval, which we did through the derivative \(f'(x)\).
• Evaluate the original function, \(f(x)\), at these critical points and at the endpoints of the interval. This ensures that potential maxima or minima aren’t missed.
• Compare these function values. The largest value is the absolute maximum, and the smallest is the absolute minimum.

In the solution, after evaluating \(f(x)\) at \(x = 0, 1,\) and \(3\), the absolute maximum was found at \(x = 1\) with \(f(1) = 1\), and the absolute minimum at \(x = 0\) with \(f(0) = 0\). This process is crucial for understanding how the function operates within the given parameters.