Problem 10 - Contemporary Precalculus

Question

Between 1997 and \(2005,\) the number \(y\) of unemployed (in thousands) was approximated by $$ \begin{array}{r} y=-53.4 x^{3}+1772.33 x^{2}-18,681.32 x+69,188.1 \\ (7 \leq x \leq 15) \end{array} $$ where \(x=7\) corresponds to \(1997 .^{\\#}\) In what year was unemployment the highest?

Step-by-Step Solution

Answer

Based on the given function and analysis, the year with the highest unemployment was 2006, with approximately 2,333,250 people unemployed.

Step 1: Calculate the derivative of the function

The derivative of the function will give us information about the slope of the function at each point. When the slope is zero, the function reaches local maximum or minimum values. The function is given by: $$y=-53.4 x^{3}+1772.33 x^{2}-18,681.32 x+69,188.1$$ So, we need to calculate its derivative, \(y'\): $$y' = \frac{d}{dx} (-53.4 x^{3}+1772.33 x^{2}-18,681.32 x+69,188.1)$$ $$y' = -160.2 x^{2} + 3544.66 x - 18,681.32$$

Step 2: Find the critical points

To find the critical points, we need to solve the equation \(y'=0\): $$-160.2 x^{2} + 3544.66 x - 18,681.32 = 0$$ This is a quadratic equation, and we can use the quadratic formula to solve it: $$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$$ where \(a = -160.2\), \(b = 3544.66\), \(c = -18,681.32\) $$x = \frac{-3544.66 \pm \sqrt{(3544.66)^2 - 4(-160.2)(-18,681.32)}}{2(-160.2)}$$ $$x \approx 9.46, 12.28$$

Step 3: Check the concavity and classify the critical points

To determine if the critical points are maximum or minimum, we can analyze the concavity of the function by calculating the second derivative, \(y''\): $$y'' = \frac{d^2}{dx^2} (-160.2 x^{2} + 3544.66 x - 18,681.32)$$ $$y'' = -320.4 x + 3544.66$$ Now, substitute the critical points into the equation for the second derivative and check their signs: For \(x \approx 9.46\): $$y''(9.46) \approx -320.4(9.46) + 3544.66 < 0$$ For \(x \approx 12.28\): $$y''(12.28) \approx -320.4(12.28) + 3544.66 > 0$$ Therefore, the function has a maximum at \(x \approx 9.46\) and a minimum at \(x \approx 12.28\). Since we are looking for the highest point, we consider the maximum, \(x \approx 9.46\).

Step 4: Identify the year and unemployment number

Since \(x = 7\) corresponds to 1997, \(x \approx 9.46\) corresponds to the year closest to \(1997 + 9.46 = 2006.46\), which is \(2006\). To find the unemployment number in thousands at this point, we can plug our \(x\) value back into the original equation: $$y \approx -53.4 (9.46)^{3}+1772.33 (9.46)^{2}-18,681.32 (9.46)+69,188.1$$ $$y \approx 2333.25$$ This indicates that the highest unemployment was approximately 2,333,250 people in the year 2006.

Key Concepts

Critical Points

Critical points of a function are where its derivative equals zero or where it is undefined. These points help us find the places where the function might have a local maximum or minimum. When we examine the graph of a function, critical points are where the curve could change direction, such as at peaks or valleys.

• They are calculated by setting the derivative of the function to zero and solving for the variable.
• Critical points give us valuable information about the behavior of a function.

By analyzing these points, we can determine where the function increases, decreases, or remains constant. In practical terms, finding the critical points helps us understand when an event, such as unemployment in a certain year, might hit a maximum or minimum.

Derivative Calculus

Derivative calculus is a fundamental tool in mathematics used to study how functions change. By calculating the derivative of a function, you can find how quickly something is happening, whether it's speeding up or slowing down.

• The derivative of a function tells us the slope or rate of change at any point on that function.
• In practical applications like the original problem, it helps us find important points like maxima and minima.

The derivative is also crucial for finding the critical points, as it becomes zero or undefined at these points. To compute the derivative, we differentiate each term of the function individually. For instance, if you have a polynomial like a cubic function, you decrease the power of each term by one and multiply by the original power.

Quadratic Equations

Quadratic equations are equations of the form \( ax^2 + bx + c = 0 \), which involve a squared term. Solving these equations is key when finding critical points, as the derivatives of polynomial functions often take this form.

• To solve a quadratic equation, we often use the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• This formula provides the roots of the equation, indicating potential critical points of the original function.

Quadratic equations can have two solutions, one solution, or no real solution, depending on the value under the square root, known as the discriminant. Understanding quadratic equations allows us to decode the behavior of more complex functions, helping us identify where they reach their peak values.