Problem 6 - Calculus for Biology and Medicine

Question

\mathrm{\\{} I n ~ P r o b l e m s ~ , each function is continuous and defined on a closed interval. It therefore satisfies the assumptions of the extreme- value theorem. With the help of a graphing calculator or spreadsheet, graph each function and locate its global extrema. (Note that a function may have more than one global minimum or maximum point.) $$ f(x)=(x-1)^{2}(x+1),-2 \leq x \leq 2 $$

Step-by-Step Solution

Answer

Global maximum is 27 at \( x = -2 \); global minimum is approximately \(-\frac{32}{27}\) at \( x = -\frac{1}{3} \).

Step 1: Understanding the Problem

We are given the function \( f(x) = (x-1)^2(x+1) \) and the interval \([-2, 2]\). Our task is to find its global extrema, that is, the highest (maximum) and lowest (minimum) values of the function on this interval.

Step 2: Graph the Function

Using a graphing tool or calculator, we graph the function \( f(x) = (x-1)^2(x+1) \) over the interval \([-2, 2]\). This helps us visually identify where the graph reaches its highest and lowest points.

Step 3: Analyzing the Graph

Upon reviewing the graph, observe the crests (peaks) and troughs (valleys) within the interval. The graph's lowest points are the global minima and the highest points are the global maxima.

Step 4: Checking Endpoints and Critical Points

Evaluate \( f(x) \) at the endpoints \( x = -2 \) and \( x = 2 \), and find critical points by setting the derivative \( f'(x) \) to zero and solving for \( x \). The derivative \( f'(x) \) can be found using the product rule.

Step 5: Calculating the Derivative

For \( f(x) = (x-1)^2(x+1) \), apply the product rule to find the derivative. After calculating, \( f'(x) = 3x^2 - 2x - 1 \). Set \( f'(x) = 0 \) to find critical points.

Step 6: Solve for Critical Points

Solve \( 3x^2 - 2x - 1 = 0 \) using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). This yields the critical points \( x = 1 \) and \( x = -\frac{1}{3} \).

Step 7: Evaluate Function at Endpoints and Critical Points

Calculate \( f(-2) \), \( f(2) \), \( f(1) \), and \( f(-\frac{1}{3}) \). You get: \( f(-2) = 27 \), \( f(2) = 3 \), \( f(1) = 0 \), and \( f(-\frac{1}{3}) \approx -\frac{32}{27} \).

Step 8: Identify Global Extrema

Compare the values at endpoints and critical points: the highest value \( 27 \) at \( x = -2 \) is the global maximum, and the lowest value \(-\frac{32}{27} \) at \( x = -\frac{1}{3} \) is the global minimum.

Key Concepts

Global Extrema

Global extrema refer to the maximum and minimum values of a function over a given interval. These are the highest and lowest points you can detect on this graph between the set boundaries. For the function \( f(x) = (x-1)^2(x+1) \) within the interval \([-2, 2]\), we need to find where it reaches its global maximum and minimum.When you're working with global extrema, you often need to

• Evaluate the function at the endpoints: Sometimes, the highest or lowest value appears at the end of the interval.
• Assess critical points: These are the points where the slope of the function is zero (derivative equals zero).

For this function, by checking endpoints and critical points (which are the values where the derivative is zero), we can determine the global maximum at \(x = -2\), where the function value is 27, and the global minimum at \(x = -\frac{1}{3}\), where the function value is approximately -\(\frac{32}{27}\). This means that within this closed interval, these are the highest and lowest outputs of the function.

Critical Points

Critical points are essential in finding extrema because they often signal where the function changes direction—from increasing to decreasing or vice versa. For the function \( f(x) = (x-1)^2(x+1) \), finding critical points involves determining where the derivative \( f'(x) \) is equal to zero.Critical points can be found by:

• Taking the derivative of the function: Use differentiation rules to find \( f'(x) \).
• Setting the derivative to zero: Solve \( f'(x) = 0 \) to find possible critical points.

For \( f(x) \), applying the product rule gives us the derivative \( f'(x) = 3x^2 - 2x - 1 \). Solving \( 3x^2 - 2x - 1 = 0 \) provides critical points at \( x = 1 \) and \( x = -\frac{1}{3}\). These critical points, along with the endpoints of the interval, are checked for potential global minima and maxima.

Product Rule

The product rule is a fundamental technique used in calculus for differentiating products of two functions. For functions like \( f(x) = (x-1)^2(x+1) \), which can be viewed as a product of \( g(x) = (x-1)^2 \) and \( h(x) = (x+1) \), the product rule helps find the derivative efficiently.The product rule states: \[(f \cdot g)' = f' \cdot g + f \cdot g' \]This means you take the derivative of the first function and multiply it by the second function, then add the product of the first function and the derivative of the second function.Applying this to \( f(x) = (x-1)^2(x+1) \):

• Derivative of \( g(x) = (x-1)^2 \) is \( 2(x-1) \).
• Derivative of \( h(x) = (x+1) \) is \( 1 \).

The product rule gives us \( f'(x) = (2(x-1))(x+1) + ((x-1)^2)(1) \), which simplifies to \( 3x^2 - 2x - 1 \). This derivative is crucial for identifying the critical points where the original function might have peaks or valleys, aiding in finding the global extrema.