Problem 75 - 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. $$ y=x^{3}-6 x+1 $$

Step-by-Step Solution

Answer

Horizontal tangents occur at \((\sqrt{2}, -4\sqrt{2} + 1)\) and \((-\sqrt{2}, 4\sqrt{2} + 1)\).

Step 1: Find the Derivative

The function given is \( y = x^3 - 6x + 1 \). To find where the tangent line is horizontal, we need to find its derivative because the derivative tells us the slope of the tangent line at any point. Thus, we differentiate \( y \) with respect to \( x \). The derivative is \( \frac{dy}{dx} = 3x^2 - 6 \).

Step 2: Set the Derivative Equal to Zero

A horizontal tangent line implies a slope of 0. Therefore, we set the derivative equal to zero: \( 3x^2 - 6 = 0 \).

Step 3: Solve for x

To solve \( 3x^2 - 6 = 0 \), first factor out the 3: \( 3(x^2 - 2) = 0 \). By dividing both sides by 3, we get \( x^2 - 2 = 0 \). Solving for \( x \), we add 2 to both sides and take the square root to obtain \( x = \pm \sqrt{2} \).

Step 4: Find the Corresponding y Values

We need to find the \( y \)-coordinate for each \( x \) value found. Substitute \( x = \sqrt{2} \) and \( x = -\sqrt{2} \) back into the original function \( y = x^3 - 6x + 1 \). For \( x = \sqrt{2} \), \( y = (\sqrt{2})^3 - 6(\sqrt{2}) + 1 = 2\sqrt{2} - 6\sqrt{2} + 1 = -4\sqrt{2} + 1 \). For \( x = -\sqrt{2} \), \( y = (-\sqrt{2})^3 - 6(-\sqrt{2}) + 1 = -2\sqrt{2} + 6\sqrt{2} + 1 = 4\sqrt{2} + 1 \).

Step 5: Write the Points

The points on the graph where the tangent is horizontal are \( (\sqrt{2}, -4\sqrt{2} + 1) \) and \( (-\sqrt{2}, 4\sqrt{2} + 1) \).

Key Concepts

Understanding the Derivative

A derivative is essentially a mathematical tool that helps us determine the rate of change of a function. Imagine driving a car; the derivative here would be your speed. It tells you how fast or slow you're moving at any given moment. In the context of functions, the derivative gives us the slope of the tangent line at any point along the curve of the function. In the problem we've been working on, our function is \( y = x^3 - 6x + 1 \). To find where the tangent line to this function is horizontal, we need to calculate its derivative because a horizontal line has a slope of 0. Here, the derivative is \( \frac{dy}{dx} = 3x^2 - 6 \). By setting this derivative equal to zero, we can find the points on the graph where the slope is zero, meaning the tangent line is perfectly horizontal.

Solving Equations for Tangent Points

Solving equations comes into play when you need to find specific values of the variable that satisfy an equation. In our example, once we find the derivative, the next step is solving \( 3x^2 - 6 = 0 \) to find the values of \( x \) that make the slope zero. This means we're hunting for points where the tangent line is horizontal.To break it down, here is what we do:

• Factor the equation: \( 3(x^2 - 2) = 0 \)
• Divide by 3: \( x^2 - 2 = 0 \)
• Add 2 to both sides and solve: \( x = \pm \sqrt{2} \)

Finding \( x = \pm \sqrt{2} \) gives specific \( x \)-values where these horizontal tangents occur. It shows us the places on the graph where the derivative is zero and therefore horizontal.

The Role of Graphing Functions

Graphing functions allows us to visually comprehend the behavior of mathematical equations. When you graph a function, you can see patterns, shapes, and transformations that are not always obvious from the algebraic form. In our case, plotting \( y = x^3 - 6x + 1 \) helps us understand where horizontal tangent lines occur. Once you know the \( x \)-values that give a horizontal slope, you can substitute these back into the function to find the corresponding \( y \)-values. For instance:

• Substitute \( x = \sqrt{2} \) to get \( y = -4\sqrt{2} + 1 \)
• Substitute \( x = -\sqrt{2} \) to get \( y = 4\sqrt{2} + 1 \)

With these coordinates, we have complete points — \( (\sqrt{2}, -4\sqrt{2} + 1) \) and \( (-\sqrt{2}, 4\sqrt{2} + 1) \) — that we can plot to see exactly where the curve has horizontal tangent lines.