Problem 49 - Calculus - AP Edition

Question

Use a graphing utility to make rough estimates of the locations of all horizontal tangent lines, and then find their exact locations by differentiating. $$y=\frac{1}{3} x^{3}-\frac{3}{2} x^{2}+2 x$$

Step-by-Step Solution

Answer

Horizontal tangents at \((1, \frac{5}{6})\) and \((2, \frac{2}{3})\).

Step 1: Understand the Problem

We need to find the locations where the function \( y = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 2x \) has horizontal tangent lines. This occurs where the derivative of the function is equal to zero.

Step 2: Differentiate the Function

Find the derivative of the function to determine where the tangent line is horizontal:\[ y' = \frac{d}{dx} \left( \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 2x \right) \]Using the power rule, the derivative is:\[ y' = x^2 - 3x + 2 \]

Step 3: Set the Derivative to Zero

To find the horizontal tangent lines, set the derivative equal to zero:\[ x^2 - 3x + 2 = 0 \]Solve this quadratic equation to find the values of \( x \).

Step 4: Solve the Quadratic Equation

Factor the quadratic equation:\[ (x - 1)(x - 2) = 0 \]Thus, the solutions are:\[ x = 1 \quad \text{and} \quad x = 2 \]

Step 5: Verify Solutions and Find \( y \) Values

Substitute \( x = 1 \) and \( x = 2 \) back into the original equation to find \( y \) values:- For \( x = 1 \): \[ y = \frac{1}{3} (1)^3 - \frac{3}{2} (1)^2 + 2(1) = \frac{1}{3} - \frac{3}{2} + 2 = \frac{5}{6} \]- For \( x = 2 \): \[ y = \frac{1}{3} (2)^3 - \frac{3}{2} (2)^2 + 2(2) = \frac{8}{3} - 6 + 4 = \frac{2}{3} \]Thus, the horizontal tangent lines are at \((1, \frac{5}{6})\) and \((2, \frac{2}{3})\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function is changing at any given point. This process is called taking the derivative of the function. When we differentiate a function, we use rules like the power rule, product rule, and chain rule, to simplify the process. For polynomial functions such as our example: \[ y = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 2x \] we apply the power rule, which states that the derivative of \( x^n \) is \( nx^{n-1} \). Using this rule for each term:

• The derivative of \( \frac{1}{3}x^3 \) becomes \( x^2 \).
• The derivative of \( -\frac{3}{2}x^2 \) becomes \( -3x \).
• The derivative of \( 2x \) becomes \( 2 \).

So, the derivative \( y' \) of our function is \( y' = x^2 - 3x + 2 \). Differentiating helps us find critical points where the slope of the tangent line is zero, which is crucial for identifying horizontal tangent lines.

Quadratic Equation

A quadratic equation is a polynomial equation of the form \( ax^2 + bx + c = 0 \). In our problem, after differentiating, we obtained the quadratic: \[ x^2 - 3x + 2 = 0 \]Solving quadratic equations involves finding the values of \( x \) that make the equation true, which can be done through various methods such as factoring, completing the square, or using the quadratic formula. In this case, the quadratic equation can be factored easily:

• The equation \( (x - 1)(x - 2) = 0 \) can be factored by identifying the roots or solutions.
• When factored, \( x - 1 = 0 \) gives us \( x = 1 \).
• And \( x - 2 = 0 \) gives us \( x = 2 \).

The solutions \( x = 1 \) and \( x = 2 \) are where the function has horizontal tangent lines. Understanding how to solve quadratic equations is essential for finding key points in calculus problems related to slopes and tangents.

Graphing Utility

A graphing utility is a valuable tool for visualizing the behavior of functions. It helps estimate the location of features like horizontal tangent lines before performing exact calculations. By entering the function \[ y = \frac{1}{3} x^{3} - \frac{3}{2} x^{2} + 2x \]into the graphing utility, we can view the curve and identify areas where the slope appears to level off, indicating horizontal tangents.

• These points typically occur when the graph changes direction, such as a peak or a valley.
• The graphing utility provides rough estimates, which are then confirmed by precise mathematical computation.

Using both visualization and calculation provides a more comprehensive understanding of the function's behavior. Graphing utilities are indispensable, as they give both students and professionals a better grasp of complex equations and their real-world implications.