Problem 7 - Essential Calculus Early Transcendentals

Question

Use the guidelines of this section to sketch the curve. $$y=\frac{1}{5} x^{5}-\frac{8}{3} x^{3}+16 x$$

Step-by-Step Solution

Answer

Critical points at \( x = \pm 2 \), local max at \( x = -2 \), min at \( x = 2 \). Concave down in \((-\infty, -2) \) and \((0, 2)\); concave up in \((-2, 0) \) and \((2, \infty)\).

Step 1: Determine the Domain

Since the function is a polynomial, its domain is all real numbers. We represent the domain as \( x \in (-\infty, \infty) \).

Step 2: Find the Derivative

To find critical points and analyze the curve, find the first derivative of the function \( y = \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x \). The derivative is: \[ y' = x^4 - 8x^2 + 16 \]

Step 3: Solve for Critical Points

Set the first derivative equal to zero to find critical points: \[ x^4 - 8x^2 + 16 = 0 \] Let \( u = x^2 \), then the equation becomes \( u^2 - 8u + 16 = 0 \). Solve by factoring or using the quadratic formula: \[ (x^2 - 4)^2 = 0 \] Thus, \( x^2 = 4 \), giving \( x = \pm 2 \).

Step 4: Second Derivative for Concavity

Find the second derivative to determine concavity: \[ y'' = 4x^3 - 16x \] To find inflection points, set \( y'' = 0 \): \[ 4x(x^2 - 4) = 0 \] The solutions are \( x = 0, \pm 2 \). Use these points to determine concavity by evaluating the sign of \( y'' \) in intervals around these points.

Step 5: Analyze Intervals for Concavity

Test intervals around the inflection points: 1. Interval \((-\infty, -2)\): Choose \( x = -3 \) then \( y'' = 4(-3)((-3)^2 - 4) = -72 \), concave down.2. Interval \((-2, 0)\): Choose \( x = -1 \) then \( y'' = 4(-1)((-1)^2 - 4) = 12 \), concave up.3. Interval \((0, 2)\): Choose \( x = 1 \) then \( y'' = 4(1)(1^2 - 4) = -12 \), concave down.4. Interval \((2, \infty)\): Choose \( x = 3 \) then \( y'' = 4(3)(3^2 - 4) = 72 \), concave up.

Step 6: Determine Local Extrema

Use the first derivative test around critical points \( x = \pm 2 \).1. For \( x = -2 \): - From left: Choose \( x = -3 \), \( y' > 0 \) (increasing). - At \( x = -2 \): \( y' = 0 \). - From right: Choose \( x = -1 \), \( y' < 0 \) (decreasing). - Local maximum at \( x = -2 \).2. For \( x = 2 \): - From left: Choose \( x = 1 \), \( y' < 0 \) (decreasing). - At \( x = 2 \): \( y' = 0 \). - From right: Choose \( x = 3 \), \( y' > 0 \) (increasing). - Local minimum at \( x = 2 \).

Step 7: Sketch the Curve

Plot the critical points \((\pm 2, y(\pm 2))\), and note local maximum and minimum. Include intervals of concavity by sketching concave down behavior in \((-\infty, -2)\) and \((0, 2)\), and concave up behavior in \((-2, 0)\) and \((2, \infty)\). The overall curve behavior is a polynomial rising to positive infinity as \( x \to \pm \infty \).

Key Concepts

Critical Points

Critical points are the values of \(x\) where a function's derivative is zero or undefined. They indicate where the function might have a local maximum, local minimum, or inflection point. For our polynomial function \(y = \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x\), we find the critical points by solving the derivative \(y' = x^4 - 8x^2 + 16\) equal to zero. This reduces to \((x^2 - 4)^2 = 0\), leading to \(x = \pm 2\). These points are significant because they help us understand at which values of \(x\) the nature of the function changes, suggesting potential peaks or valleys in the curve. Recognizing these points is crucial in curve sketching as it allows us to identify the main features of the graph.

Derivative Analysis

Derivative analysis involves calculating the first and second derivatives of a function to evaluate its behavior. The first derivative, \(y' = x^4 - 8x^2 + 16\), helps us find critical points where the slope of the tangent to the curve is zero. These are locations on the graph where the curve may have extreme points—either maxima or minima. The second derivative, \(y'' = 4x^3 - 16x\), is used to determine concavity and inflection points. By examining the sign changes of the first derivative around critical points, and using the second derivative, we gain insight into the increasing or decreasing nature of the function and its curvature behavior, like where the graph bends upwards or downwards.

Concavity

Concavity refers to the direction in which a curve opens or bends. This is determined by the sign of the second derivative \(y''\). If \(y'' > 0\), the function is concave up (like a cup), and if \(y'' < 0\), it is concave down (like a cap). For the example \((x^2 - 4)^2 = 0\), leading to critical points at \(x = 0, \pm 2\). Then, we analyze intervals defined by these points:

• \((-\infty, -2)\): Concave down, since \(y'' < 0\).
• \((-2, 0)\): Concave up, since \(y'' > 0\).
• \((0, 2)\): Concave down, since \(y'' < 0\).
• \((2, \infty)\): Concave up, since \(y'' > 0\).

Concavity gives us detailed understanding of the graph's shape, allowing accurate sketching of the curve's upswings and downswings.

Polynomial Functions

Polynomial functions are expressions involving variables raised to whole number powers, with coefficients. These functions are smooth and continuous over real numbers. The provided function, \(y = \frac{1}{5}x^5 - \frac{8}{3}x^3 + 16x\), is an example of a polynomial of degree 5, a quintic function. Such functions can have multiple bends and turns due to their higher degree. Polynomial functions tend to rise to positive or negative infinity as \(x\) goes to positive or negative infinity, respectively. The domain of polynomial functions is all real numbers, \(x \in (-\infty, \infty)\). Analyzing these functions includes getting derivatives to understand the behavior and identify key features, such as critical points and intervals where they increase or decrease.