Problem 39 - Calculus of a Single Variable

Question

Finding Extrema on an Interval In Exercises \(37-40\) , find the absolute extrema of the function (if any exist) on each interval. $$ \begin{array}{l}{f(x)=x^{2}-2 x} \\ {\begin{array}{ll}{\text { (a) }[-1,2]} & {\text { (b) }(1,3]} \\ {\text { (c) }(0,2)} & {\text { (d) }[1,4)}\end{array}}\end{array} $$

Step-by-Step Solution

Answer

The absolute extrema of the function \(f(x) = x^{2} - 2x\) on the intervals are (Max, Min): (a) [-1,2] = (1, -1), (b) [1,3] = (5, 1), (c) [0,2] = (2, 0), and (d) [1,4) = (12, 1).

Step 1: Find the derivative of the function

The derivative of the function \(f(x) = x^{2} - 2x\) is \(f'(x) = 2x - 2\).

Step 2: Set the derivative equal to zero

We find the critical points by setting the derivative equal to zero thus, \(2x - 2 = 0\). Solving for \(x\) gives \(x = 1\). So, \(x = 1\) is the critical point of the function.

Step 3: Find function values at endpoints and critical point

Now evaluate the function at the endpoints and the critical point for each interval. For example, for the interval [-1,2], compute \(f(-1)\), \(f(1)\), and \(f(2)\). Whichever yields the highest result is the maximum, while the lowest result is the minimum.

Step 4: Repeat for each interval

Repeat Step 3 for each of the given intervals [(a) [-1,2], (b) [1,3], (c) [0,2], (d) [1,4)]. For each interval, compare the values obtained to determine the minimum and maximum.

Key Concepts

Understanding Critical Points

When dealing with functions, especially when identifying extrema, critical points become essential. A critical point of a function is where the function's derivative is zero or undefined. For the function in our exercise, where \( f(x) = x^2 - 2x \), its derivative \( f'(x) = 2x - 2 \) is zero at \( x = 1 \). This means that \( x = 1 \) is a critical point.

Why are critical points important? Critical points are the candidates for local maxima or minima, but not all critical points guarantee an extremum. That’s why we evaluate the function at these points and the endpoints of an interval to find absolute extrema.

• Set the derivative equal to zero.
• Solve for \( x \) to find potential extremes within an interval.
• Remember that critical points could correspond to maxima, minima, or neither.

Derivative of a Function

The derivative of a function gives us the rate at which the function's value changes as \( x \) changes. It can be thought of as the slope of the tangent line to the function's graph at any point.

For our exercise, finding the derivative, \( f'(x) \), was straightforward. We began with \( f(x) = x^2 - 2x \). Taking the derivative using the power rule, where the derivative of \( x^n \) is \( nx^{n-1} \), resulted in \( f'(x) = 2x - 2 \).

• The derivative is calculated to understand how a function increases or decreases.
• Zeros of the derivative indicate possible changes in the direction of the function.
• Derivative evaluation helps pinpoint critical points, a key step in finding extrema.

Using Interval Notation

Interval notation is a shorthand way of writing the set of points that lie between two endpoints. It can seem like a small detail, but correctly interpreting this notation is crucial when solving problems about extrema.

In our exercise, intervals are provided like \([-1,2]\) or \((1,3]\). Square brackets \([ \text{or} ]\) indicate that an endpoint is included in the interval, also known as 'closed', while parentheses \(( \text{or} )\) mean that the endpoint is not included, known as 'open'.

• Closed interval (e.g., \([-1,2]\)) includes the endpoints.
• Open interval (e.g., \((1,3]\)) does not include the starting point but includes the ending point.
• Mixing open and closed notations affects which values are checked when evaluating endpoints.

Function Evaluation at Points

Function evaluation is about finding the output of a function at specific inputs. This step is pivotal when determining the extrema by assessing how a function behaves at endpoints and critical points.

For each interval, such as \([-1,2]\), it involves computing values like \( f(-1) \), \( f(1) \), and \( f(2) \). By substituting these values into \( f(x) = x^2 - 2x \), we find specific outputs.

• Calculate function values at each interval's boundaries and critical points.
• The highest value in the interval provides the maximum.
• The lowest value gives the minimum.
• This process identifies absolute extrema within a given interval.