Problem 76 - Calculus and its applications

Question

For each function, find the points on the graph at which the tangent line is horizontal. If none exist, state that fact. $$ y=\frac{1}{3} x^{3}-3 x+2 $$

Step-by-Step Solution

Answer

Tangent lines are horizontal at \((\sqrt{3}, -2\sqrt{3} + 2)\) and \((-\sqrt{3}, 2\sqrt{3} + 2)\).

Step 1: Understand What Horizontal Tangent Means

The tangent of a function is horizontal where the derivative of the function is equal to zero. Therefore, to find where the tangent is horizontal, we first need to compute the derivative of the given function.

Step 2: Differentiate the Function

Given the function \( y = \frac{1}{3}x^3 - 3x + 2 \), we find the derivative with respect to \( x \). The derivative of \( \frac{1}{3}x^3 \) is \( x^2 \), the derivative of \( -3x \) is \( -3 \), and the derivative of a constant \( 2 \) is zero. Therefore, \( \frac{dy}{dx} = x^2 - 3 \).

Step 3: Set the Derivative Equal to Zero

To find where the tangent line is horizontal, set \( \frac{dy}{dx} = 0 \). This gives the equation \( x^2 - 3 = 0 \). Solve for \( x \) to find the points of interest.

Step 4: Solve for x

Solve the equation \( x^2 - 3 = 0 \) by adding 3 to both sides to get \( x^2 = 3 \). Take the square root of both sides to find \( x = \pm \sqrt{3} \).

Step 5: Find Corresponding y-values

Substitute \( x = \sqrt{3} \) and \( x = -\sqrt{3} \) back into the original function to find the corresponding \( y \)-values. For \( x = \sqrt{3} \), \( y = \frac{1}{3}(\sqrt{3})^3 - 3\sqrt{3} + 2 \). For \( x = -\sqrt{3} \), compute in a similar manner. Simplify both to obtain the exact \( y \)-values.

Step 6: Calculate and Simplify y-values

The exact points are found by calculating the \( y \)-coordinates: For \( x = \sqrt{3} \), \( y = \sqrt{3} - 3\sqrt{3} + 2 = -2\sqrt{3} + 2 \). For \( x = -\sqrt{3} \), \( y = -\sqrt{3} + 3\sqrt{3} + 2 = 2\sqrt{3} + 2 \). Therefore, the points are \( (\sqrt{3}, -2\sqrt{3} + 2) \) and \( (-\sqrt{3}, 2\sqrt{3} + 2) \).

Key Concepts

Understanding Derivative Calculation

When we want to determine the points on a curve where the tangent line is horizontal, the key step is to calculate the derivative of the function. The derivative tells us the slope of the tangent line at any given point on the curve.
It reveals how the function changes at different rates and helps in identifying any stationary points where the slope is zero.
To find the derivative of the function \( y = \frac{1}{3}x^3 - 3x + 2 \):

• The term \( \frac{1}{3}x^3 \) differentiates to \( x^2 \), following the power rule \( nx^{n-1} \).
• The term \( -3x \) becomes \( -3 \).
• The constant term \( 2 \) differentiates to \( 0 \), since the derivative of a constant is always zero.

Thus, the derivative of the function is \( \frac{dy}{dx} = x^2 - 3 \). Setting this derivative equal to zero will allow us to find where the tangent is horizontal.

Exploring the Function Graph

The graph of a function gives us a visual representation of how the function behaves over different values of \(x\). In particular, it helps us see where the slope of the tangent line is horizontal, meaning the derivative is zero.
When looking at the function \( y = \frac{1}{3}x^3 - 3x + 2 \), we can observe certain characteristics:

• The graph is a cubic polynomial, which means it has a characteristic 'S' shape, indicating it will turn around at least twice.
• The points where the derivative becomes zero will be the peaks and troughs, which represent local maxima and minima respectively.

Understanding the layout of the graph helps us connect our calculations to a visual component, aiding in comprehending where exactly on the graph these horizontal tangents occur.

Techniques for Solving Equations

Solving the equation \( x^2 - 3 = 0 \) involves algebraic manipulation to find the values of \( x \) where the tangent is horizontal.
Here's a simple procedure to solve it:

• First, add 3 to both sides of the equation to isolate the \( x^2 \) term, yielding \( x^2 = 3 \).
• Next, take the square root of both sides of the equation to solve for \( x \). Remember, taking a square root results in two solutions: \( x = \sqrt{3} \) and \( x = -\sqrt{3} \).

By substituting these \( x \)-values back into the original function, we find the corresponding \( y \)-values:

• For \( x = \sqrt{3} \), \( y = \sqrt{3} - 3\sqrt{3} + 2 = -2\sqrt{3} + 2 \).
• For \( x = -\sqrt{3} \), \( y = -\sqrt{3} + 3\sqrt{3} + 2 = 2\sqrt{3} + 2 \).

Thus, the solutions provide the exact points where the tangent line to the function is horizontal.