Problem 40 - Essential Calculus Early Transcendentals

Question

At which points on the curve \(y=1+40 x^{3}-3 x^{5}\) does the tangent line have the largest slope?

Step-by-Step Solution

Answer

The tangent line has the largest slope at \(x = \pm 2\sqrt{2}\).

Step 1: Find the Derivative

To determine where the tangent line has the largest slope, we first need to find the derivative of the function \(y = 1 + 40x^3 - 3x^5\). The derivative of a function gives us the slope of the tangent line at any point \(x\). Using the power rule, differentiate term by term:\[\frac{dy}{dx} = 0 + 120x^2 - 15x^4 = 120x^2 - 15x^4.\]

Step 2: Find Critical Points

To find the points where the slope reaches its maximum, we need to find the critical points by setting the derivative equal to zero:\[120x^2 - 15x^4 = 0.\]Factor out the greatest common factor:\[15x^2(8 - x^2) = 0.\]This gives us two possible solutions:\[15x^2 = 0 \quad \text{or} \quad 8 - x^2 = 0.\]So, \(x = 0\) or \(x = \pm \sqrt{8} = \pm 2\sqrt{2}\).

Step 3: Determine Maximum Slope

To ensure that these critical points correspond to maximum slopes, we analyze the second derivative to check the concavity at these points. The second derivative is:\[\frac{d^2y}{dx^2} = \frac{d}{dx}(120x^2 - 15x^4) = 240x - 60x^3.\]

Step 4: Second Derivative Test

Evaluate the second derivative at the critical points to determine if they correspond to a maximum slope:1. For \(x = 0\): \[\frac{d^2y}{dx^2} = 240(0) - 60(0)^3 = 0,\] which is inconclusive.2. For \(x = 2\sqrt{2}\): \[\frac{d^2y}{dx^2} = 240(2\sqrt{2}) - 60(2\sqrt{2})^3 = 480\sqrt{2} - 960\sqrt{2} = -480\sqrt{2},\] which is negative, indicating a local maximum.3. For \(x = -2\sqrt{2}\): The calculation is similar to \(x = 2\sqrt{2}\) and also results in a negative value, indicating a local maximum.

Key Concepts

Derivative

A derivative is a fundamental concept in calculus that measures how a function changes as its input changes. It is often represented as \( \frac{dy}{dx} \), meaning the rate of change of \( y \) with respect to \( x \). The derivative provides the slope of the tangent line to the curve at any given point.

To find the derivative of \( y = 1 + 40x^3 - 3x^5 \), we use the power rule. The power rule states that for any term \( ax^n \), the derivative is \( anx^{n-1} \).

Applying this:

• For 1, the derivative is 0.
• For \( 40x^3 \), the derivative is \( 120x^2 \).
• For \( -3x^5 \), the derivative is \( -15x^4 \).

Combine the results to get the derivative: \( \frac{dy}{dx} = 120x^2 - 15x^4 \).

This function describes how steeply the curve rises or falls at any point \( x \). Finding where this derivative is zero will help us determine critical points where the slope might achieve a maximum or minimum.

Tangent Line

The tangent line to a curve at a given point is a straight line that just "touches" the curve at that point. It has the same slope as the curve itself does at that specific point. In practical terms, the tangent line represents the immediate direction in which the curve is heading.

For the function \( y = 1 + 40x^3 - 3x^5 \), the slope of the tangent line at any point \( x \) can be found by evaluating the derivative \( 120x^2 - 15x^4 \).

This value indicates exactly how much \( y \) increases or decreases as \( x \) moves a tiny bit from its current position. The line with this slope looks like the function does, but locally and for infinitesimal movements around \( x \).

Determining where the tangent line has the largest slope involves finding when the derivative reaches its largest positive or smallest negative values, which typically happens at critical points.

Critical Points

Critical points on a graph occur where the derivative is zero or undefined. They indicate where the function can have a potential maximum, minimum, or point of inflection.

To find these, set the derivative equal to zero and solve for \( x \): \( 120x^2 - 15x^4 = 0 \).

Solve this equation by factoring:

• Factor out the greatest common factor: \( 15x^2(8 - x^2) = 0 \).
• This gives solutions: \( 15x^2 = 0 \) and \( 8 - x^2 = 0 \).
• Thus, \( x = 0 \) and \( x = \pm 2\sqrt{2} \).

These solutions are the critical points. They are where the slope is zero, and we should check them further to see if they correspond to a local maximum or minimum slope using the second derivative test.

Second Derivative Test

The second derivative test is a method used to determine whether a given critical point is a local maximum, minimum, or a point of inflection by assessing the concavity of the function at that point.

Evaluate the second derivative \( \frac{d^2y}{dx^2} = 240x - 60x^3 \) at each critical point:

• For \( x = 0 \): \( \frac{d^2y}{dx^2} = 240(0) - 60(0)^3 = 0 \). Here, the test is inconclusive.
• For \( x = 2\sqrt{2} \) and \( x = -2\sqrt{2} \):
  • Calculate \( \frac{d^2y}{dx^2} = 480\sqrt{2} - 960\sqrt{2} = -480\sqrt{2} \).
  • The negative result indicates concavity down, meaning these critical points are indeed local maxima.

By using the second derivative test, it's confirmed that the tangent line has its largest slopes at \( x = \pm 2\sqrt{2} \) on this curve, as the derivative transitions from positive to negative slope.