Problem 46 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\left|x^{2}-2 x\right|$$

Step-by-Step Solution

Answer

Local minimum at (1, -1). No inflection points. Graph transitions at x = 0 and x = 2.

Step 1: Rewrite the Absolute Value Function

The function given is expressed with an absolute value. To handle this, we break it into two cases based on the expression inside the absolute value: 1. **If** \(x^2 - 2x \geq 0\), then \(y = x^2 - 2x\). 2. **If** \(x^2 - 2x < 0\), then \(y = -(x^2 - 2x) = 2x - x^2\).

Step 2: Solve for Critical Points

To find critical points, compute the derivative of both cases. **Case 1:** \(y = x^2 - 2x\) - Derivative: \(y' = 2x - 2\)- Set \(y' = 0\): \(2x - 2 = 0\) gives \(x = 1\)**Case 2:** \(y = 2x - x^2\) - Derivative: \(y' = 2 - 2x\)- Set \(y' = 0\): \(2 - 2x = 0\) gives \(x = 1\)The critical point in both cases is \(x = 1\). Evaluate at this point for extremum candidates.

Step 3: Evaluate Second Derivative for Concavity

**Case 1:** - Second Derivative: \(y'' = 2\), which is positive, indicating a concave up shape in this case (no inflection points). **Case 2:** - Second Derivative: \(y'' = -2\), which is negative, indicating a concave down shape in this case (no inflection points).

Step 4: Analyze the Absolute Value and Extrema

Determine changes caused by the absolute value:- The points where \(x^2 - 2x = 0\) are where the sign changes. Solve: \(x(x-2) = 0\) gives \(x = 0\) and \(x = 2\).- Check the value at \(x = 0\) and \(x = 2\). Both give \(y = 0\), creating potential transition points.

Step 5: Identify Local and Absolute Extrema

Using the behavior of the function across critical points and transition points:- **Local Extremum:** At \(x = 1\), evaluate in both cases: \(y = 1 - 2 = -1\). Thus, \((1, -1)\) is a local minimum given it is the lowest value from comparison.- **Absolute Extremum:** The function's minimum value is achieved at the local minimum \((1, -1)\). Absolute maximum is undefined since it increases infinitely.

Step 6: Graph the Function

Graph consists of segments: - From \(x = -fty\) to \(x = 0\), the function follows the expression \(y = 2x - x^2\).- From \(x = 0\) to \(x = 2\), it follows \(y = x^2 - 2x\).- Beyond \(x = 2\), follows \(y = x^2 - 2x\) again and the values of \(y\) smoothly pass through this transition.

Key Concepts

Critical Points Identification

Identifying critical points is crucial for understanding where a function might have a local maximum, minimum, or potential saddle points. Critical points occur where the derivative of the function is zero or does not exist.

• First, take the derivative of the function. For the two cases of our absolute value function: - When \(y = x^2 - 2x\), the derivative \(y' = 2x - 2\), and - When \(y = 2x - x^2\), it is \(y' = 2 - 2x\).
• Set each derivative to zero to solve for critical points. For both cases, this leads to \(x = 1\).

To determine if these are indeed critical points of the original function with absolute values, check also if the sign changes at these x-values within the context of the piecewise function. Knowing your critical points helps you prepare for the next step in finding extrema using subsequent tests.

Absolute Value Functions

The challenge of handling absolute value functions stems from their non-linear, piecewise nature. Here, the function \(y = \left| x^2 - 2x \right| \) is broken into separate expressions depending on whether \(x^2 - 2x\) is positive or negative.

• When \(x^2 - 2x \geq 0\), the function is simply \(y = x^2 - 2x\).
• When \(x^2 - 2x < 0\), you must flip the sign to get \(y = 2x - x^2\).

Identifying the regions where each function applies involves solving \(x^2 - 2x = 0\). This gives the transition points \(x = 0\) and \(x = 2\), which are crucial for drawing the overall graph of the function and understanding where the absolute value affects the curve.

Second Derivative Test

The second derivative test helps to determine the concavity of a function around a critical point, which in turn reveals whether that point is a local maximum, minimum, or a point of inflection (though note that true inflection points require the first derivative to be changing sign as well).

• Calculate the second derivatives for both cases: - For \(y = x^2 - 2x\), you get \(y'' = 2\), which is positive, indicating the graph is concave up near this region. - For \(y = 2x - x^2\), the second derivative is \(y'' = -2\), which is negative, indicating a concave down shape.

Thus, concavity at \(x = 1\) confirms that this critical point is indeed a local minimum for both expressions. This test effectively complements the critical points identification step, reinforcing our conclusions about the extreme points of the function.