Problem 45 - Thomas Calculus in SI Units

Question

Identify the coordinates of any local and absolute extreme points and inflection points. Graph the function. $$y=\left|x^{2}-1\right|$$

Step-by-Step Solution

Answer

Local maxima at (0, 1); local and absolute minima at (-1, 0) and (1, 0); no inflection points.

Step 1: Identify the Piecewise Function

The function is given as an absolute value, which means it can be expressed as a piece-wise function. For \(y = |x^2 - 1|\), we have two cases: 1. If \(x^2 - 1 \ge 0\), then \(y = x^2 - 1\).2. If \(x^2 - 1 < 0\), then \(y = -(x^2 - 1) = 1 - x^2\). The expression \(x^2 - 1 = 0\) can be solved as \(x = \pm 1\), therefore, the intervals are \((-\infty, -1]\), \((-1, 1)\), and \([1, \infty)\).

Step 2: Find Critical Points and Analyze Local Extrema

Each piece of the function can have its derivative calculated separately.For \(y = x^2 - 1\) (when \(x \le -1\) or \(x \ge 1\)), the derivative is \(y' = 2x\).For \(y = 1 - x^2\) (when \(-1 < x < 1\)), the derivative is \(y' = -2x\).Setting \(y' = 0\) in each case gives the critical points. For \(-1 < x < 1\), \(-2x = 0\) implies \(x = 0\), thus, there is a critical point at \((0, 1)\).Check at boundaries: \((1, 0)\) and \((-1, 0)\).

Step 3: Evaluate Critical Points for Local and Absolute Extrema

Evaluate the function at critical points and boundaries:- At \(x = 0\), \(y(0) = |0^2 - 1| = 1\).- At \(x = -1\), \(y(-1) = |-1^2 - 1| = 0\).- At \(x = 1\), \(y(1) = |1^2 - 1| = 0\).Since \((0, 1)\) is higher than both \((-1, 0)\) and \((1, 0)\), \((0, 1)\) is a local maximum. \((-1, 0)\) and \((1, 0)\) are local minima. Since \(y\) can't go lower than 0, they are also absolute minima.

Step 4: Determine Inflection Points

Analyze the concavity by calculating the second derivative. For \(y = x^2 - 1\), second derivative is \(y'' = 2\), indicating concave up for \(x \le -1\) and \(x \ge 1\).For \(y = 1 - x^2\), second derivative is \(y'' = -2\), indicating concave down for \(-1 < x < 1\).The change in concavity occurs at the transition points \(x = -1\) and \(x = 1\), but the function's absolute value form does not create differentiability at these transitions, thus there are no inflection points.

Step 5: Graph the Function

Plot the function based on analyzed behavior:- From \(x = -\infty\) to \(x = -1\), the graph follows \(y = x^2 - 1\) and decreases to \(y = 0\) at \(x = -1\).- From \(x = -1\) to \(x = 1\), the graph follows \(y = 1 - x^2\) forming an upside down parabola peaking at \(x = 0\), \(y = 1\).- From \(x = 1\) to \(x = \infty\), the graph follows \(y = x^2 - 1\) and increases from \(y = 0\) at \(x = 1\). The transitions at \(x = -1\) and \(x = 1\) align smoothly. Note changes in direction at these points.

Key Concepts

Piecewise Function

A piecewise function is a function that has different expressions for different segments of its domain. In the case of the function \(y = |x^2 - 1|\), the absolute value indicates that our function will split into two cases depending on the value inside the absolute value function. This can be interpreted as:

• \( y = x^2 - 1 \) when \( x^2 - 1 \ge 0 \), which simplifies to x-values \((-\infty, -1]\) and \([1, \infty)\).
• \( y = -(x^2 - 1) = 1 - x^2 \) when \( x^2 - 1 < 0 \), applicable for x-values \((-1, 1)\).

The splitting points are determined by solving \(x^2 - 1 = 0\), giving us \(x = \pm 1\). This creates intervals where the function behaves differently, forming a piecewise function.

Absolute Value Function

Absolute value functions like \(f(x) = |x|\) take any real number and return its non-negative counterpart. This property translates our function \(y = |x^2 - 1|\) into non-negative outputs regardless of whether \(x^2 - 1\) is positive or negative. The absolute value function is crucial in determining the piecewise nature of the function. Here's how it works:

• When \(x^2 - 1 \ge 0\), the output is simply \(x^2 - 1\).
• When \(x^2 - 1 < 0\), the output flips to \(-(x^2 - 1)\) which equals \(1 - x^2\). This ensures all outputs are non-negative.

This transformation highlights the significance of absolute values in creating functions with non-negative outputs, while providing a continuous transition over differing inputs.

Critical Points

Critical points in calculus are where the derivative of a function is zero or undefined, indicating potential local extrema or transitions. To find the critical points of our piecewise function, we separate the function's derivative analysis based on its defined parts:

• For \(y = x^2 - 1\), the derivative is \(y' = 2x\). There are no zero derivative points within \([-\infty, -1)\cup (1, \infty)\).
• For \(y = 1 - x^2\), the derivative is \(y' = -2x\). Solving \(-2x = 0\), we find a critical point at \(x = 0\) in the interval \((-1, 1)\).

Additionally, the boundary values where the function changes its piece—\(x = -1\) and \(x = 1\)—are also checked, revealing points like \((0, 1)\) as local maxima and \((-1, 0)\) and \((1, 0)\) as local minima.

Concavity Analysis

Concavity helps describe the curvature of a function. It is determined by the sign of the second derivative. For our function:

• For \(y = x^2 - 1\), the second derivative is \(y'' = 2\), indicating a concave up profile over \(x \le -1\) and \(x \ge 1\).
• For \(y = 1 - x^2\), the second derivative is \(y'' = -2\), indicating a concave down form over \(-1 < x < 1\).

Although transitions in concavity suggest potential inflection points, the absolute value aspect of our function makes it non-differentiable at \(x = -1\) and \(x = 1\), ruling out inflection points. Thus, despite changes in concavity, no inflection points occur because the original function transitions across domains in a piecewise nature.