Problem 29 - Calculus for Biology and Medicine

Question

Determine whether the functions have absolute maxima and minima, and, if so, find their coordinates. Find inflection points. Find the intervals on which the function is increasing, on which it is decreasing, on which it is concave up, and on which it is concave down. Sketch the graph of each function. $$ y=\left|x^{2}-9\right|,-4 \leq x \leq 5 $$

Step-by-Step Solution

Answer

Absolute maximum at \((5, 16)\); minima at \((-3, 0)\) and \((3, 0)\). Inflection points at \(x = -3\) and \(x = 3\).

Step 1: Analyzing the function

We have the function \( y = |x^2 - 9| \). This function is defined as the absolute value of a quadratic. It can be expressed as a piecewise function: \( y = x^2 - 9 \) for \( x^2 \geq 9 \) (i.e., \( x \leq -3 \) or \( x \geq 3 \)), and \( y = -(x^2 - 9) = 9 - x^2 \) for \( -3 < x < 3 \). We need to analyze these pieces within the given interval \([-4, 5]\).

Step 2: Finding critical points

First, we find where the derivative of the function is zero or undefined. The derivative of \( y = x^2 - 9 \) is \( y' = 2x \), and it's zero at \( x = 0 \). The derivative of \( y = 9 - x^2 \) is \( y' = -2x \), and it's zero at \( x = 0 \). The points where the function switches between these pieces, \( x = \pm 3 \), are also critical because the derivative is undefined at these points.

Step 3: Determine increasing/decreasing intervals

Evaluate \( y' \) in intervals \([-4, -3)\), \([-3, 0)\), \((0, 3)\), and \((3, 5]\). For \( x < -3 \) and \( x > 3 \), \( y' = 2x \) is positive or negative depending on \( x > 0 \) or \( x < 0 \). For \(-3 < x < 3\), \( y' = -2x \), giving increasing for \( x < 0 \) and decreasing for \( x > 0 \). Thus the function increases on \([-4, -3) \cup (0, 3)\) and decreases on \([-3, 0) \cup (3, 5]\).

Step 4: Identify concavity and inflection points

We use the second derivative test. For \( x < -3 \) or \( x > 3 \), \( y'' = 2 \), indicating concave up. For \(-3 < x < 3\), \( y'' = -2 \), indicating concave down. Inflection points occur where \( y'' \) changes sign: at \( x = -3 \) and \( x = 3 \).

Step 5: Finding extrema

Examine the function value at critical points and endpoints: \( x = -4, -3, 0, 3, 5 \). Calculate: \( y(-4) = 7, y(-3) = 0, y(0) = 9, y(3) = 0, y(5) = 16 \). Thus, absolute minimum is \(y = 0\) at \((-3, 0)\) and \((3, 0)\), maximum is \(y = 16\) at \((5, 16)\).

Step 6: Sketching the graph

Draw the piecewise segments for \( y = -(x^2 - 9) \) between \( x = -3 \) and \( x = 3 \), showing downward curvature. The segments for \( y = x^2 - 9 \) are upward outside this interval, with the graph meeting the x-axis at \( x = \pm3 \). The graph dips to \( y = 0 \) at \( x = \pm 3 \), reaches a minimum at \( x = 0 \) intersecting when changing concavity.

Key Concepts

Absolute Maxima and Minima

To understand absolute maxima and minima, think of them as the 'tallest' and 'lowest' points a function can reach within a specific interval. For the function \( y = |x^2 - 9| \) over the range \([-4, 5]\), it is crucial to evaluate the endpoints and any critical points. At these critical points, the derivative is either zero or undefined, meaning the slope of the function is flat or changes undetermined.

For this function,

• The critical points occur at \( x = -3, 0, \text{ and } 3 \).
• At each of these points, we calculated the y-values and found that:
• The absolute maximum is \( y = 16 \) at \((5, 16)\), meaning that's the highest point on the curve in the interval given.
• The absolute minima are at \((3, 0)\) and \((-3, 0)\), where the curve touches its lowest at these points.

This step ensures that students understand how to evaluate a function to find these max and min values and not just rely on calculus tools provided by derivatives.

Inflection Points

Inflection points are where a graph changes its concavity, switching from being 'curvy' upwards to 'curvy' downwards, or vice versa. For the function \( y = |x^2 - 9| \), the inflection points occur where the second derivative transitions in its sign.

• In this problem, by analyzing the second derivative test, we learn that:
• For \( x < -3 \) and \( x > 3 \), the second derivative \( y'' = 2 \); it's positive.
• For \(-3 < x < 3\), \( y'' = -2 \), indicating the function is concave down where the second derivative is negative.
• Inflection points are exactly where this sign change occurs: at \( x = -3 \) and \( x = 3 \).

Recognizing inflection points is crucial because they tell us where the function shifts its 'bow' and might impact the graph's overall shape.

Concavity

Concavity describes how curved the graph is and which direction it curves towards. When we refer to concave up, it resembles a cup ('U' shape), and concave down looks like a frown ('∩' shape).

For the given function, noting

• For values \( x < -3 \) or \( x > 3 \), the function is concave up because \( y'' = 2 \).
• In this section of the function, it appears like a cup opening upwards.
• Inside \(-3 < x < 3\), the function is concave down as \( y'' = -2 \), meaning it turns downwards.

Understanding concavity is vital for sketching graphs accurately and interpreting the behavior of the function through different intervals.

Increasing and Decreasing Intervals

Functions often rise and fall within different sections over their domain. Identifying increasing and decreasing intervals reveals where the function is climbing or dipping.

For \( y = |x^2 - 9| \), we use the derivative to analyze:

• When \( y' > 0 \), the function is rising or increasing.
• When \( y' < 0 \), the function descends or decreases.
• Between \([-4, -3)\) and \((0, 3)\), \( y' \) indicates increasing behavior, meaning the curve climbs upwards.
• Between \([-3, 0)\) and \((3, 5]\), \( y' \) is pointing to decreasing behavior where the graph falls.

Identifying these intervals helps in sketching the graph fluidly and predicting how the function behaves throughout its range.