Problem 6 - Calculus for Biology and Medicine

Question

Find the local maxima and minima of each of the functions. Determine whether each function has local maxima and minima and find their coordinates. For each function, find the intervals on which it is increasing and the intervals on which it is decreasing. \(y=\left|x^{2}-25\right|,-5 \leq x \leq 8\)

Step-by-Step Solution

Answer

Local minima at \((-5, 0)\) and \((5, 0)\); decreasing on \((-5, 5)\); increasing on \((-5, -5)\) and \((5, 8)\). No local maxima.

Step 1: Analyze the Function

The given function is \( y = |x^2 - 25| \). This is an absolute value function, which means it can change its behavior (e.g., from increasing to decreasing) at points where the expression inside the absolute value changes sign.

Step 2: Find Critical Points

To identify critical points, consider where the expression inside the absolute value changes sign. Set \( x^2 - 25 = 0 \), which simplifies to \( x^2 = 25 \). Solving for \( x \), we get \( x = \pm 5 \). These are the critical points where the function's behavior may change.

Step 3: Evaluate the Function at Critical Points and Bounds

Calculate \( y \) at critical points and at the boundaries (-5 and 8). At \( x = -5 \), \( y = |(-5)^2 - 25| = 0 \). At \( x = 5 \), \( y = |5^2 - 25| = 0 \). At \( x = 8 \), \( y = |8^2 - 25| = 39 \).

Step 4: Determine Intervals for Increasing and Decreasing

Examine \( y = x^2 - 25 \) around critical points:- For \( x < -5 \): \( x^2 - 25 > 0 \), so function is decreasing.- For \( -5 < x < 5 \): \( x^2 - 25 < 0 \), thus function becomes increasing as the absolute reverses sign.- For \( x > 5 \): \( x^2 - 25 > 0 \), so function is again increasing.

Step 5: Identify Local Maxima and Minima

The function reaches its minimum value of 0 at \( x = -5 \) and \( x = 5 \). Since there’s an interval switch, the behavior of the absolute value causes these to be local minima. There are no local maxima within the given interval, as the endpoints are not lower than any nearby points on the function.

Key Concepts

Local Maxima

A local maximum occurs at a point where a function reaches the highest value compared to its immediate surroundings. When examining the specific function \[ y = |x^2 - 25| \]we look for places within the interval where the function could exhibit a peak before it dips down again. The function exhibits changes around the defined critical points, but in this situation, there are no local maxima because the function's behavior at these points results in local minima instead.

For a local maximum to occur, the value of the function at a certain point would need to be higher than that of its neighboring points. In this exercise, neither -5 nor 5 fulfill this condition. Both points, -5 and 5, give the function value of 0, which do not exceed the values near them. Furthermore, at the boundary point, \( x = 8 \), the function value is 39, which is not a maximum compared to any potential value before it increases again.

Local Minima

Local minima are the points where a function reaches its lowest value in a neighborhood. In the context of the function \[ y = |x^2 - 25| \]a local minimum occurs where the function value is lower than its immediately neighboring values. Such phenomenon is visible at critical points of the function.

Examining critical points, we find that the function has local minima at \( x = -5 \) and \( x = 5 \). Both these points yield the function value \[ y = 0 \].

To determine if these are indeed local minima, we observe the surrounding intervals:

• For \( x < -5 \), the function decreases towards \( x = -5 \) and begins increasing again past this point.
• For \(-5 < x < 5\), the function increases up to \( x = 5 \) before it starts decreasing past this point.

These behaviors confirm that both critical points are indeed local minima because the function value at these points is the lowest compared to its immediate neighboring points.

Increasing and Decreasing Intervals

Understanding where a function increases or decreases within its domain helps visualize its behavior over a range. For the function \[ y = |x^2 - 25| \]determining such intervals guides us in interpreting the function's trajectory.

Analyzed critical points and specific sections yield the following conclusions:

• From \( x < -5 \), the section \( x^2 - 25 > 0 \), meaning the function is decreasing.
• Between \( x = -5 \) and \( x = 5 \), the section \( x^2 - 25 < 0 \), indicating the function is increasing.
• For \( x > 5 \), the section \( x^2 - 25 > 0 \), showing that the function increases once more.

As a result, the intervals can be summarized as decreasing at \((-5, -∞)\) and increasing within the interval \((−5, 5)\). Then, starting from \( x = 5 \), the function increases again along \((5, ∞)\). Overall, recognizing these intervals helps in depicting a clear image of the function's behavior, providing insights into its peaks and troughs across the domain.