Problem 22 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 19-28, determine where the graph of the given function is increasing, decreasing, concave up, and concave down. Then sketch the graph (see Example 4). \(F(x)=x^{6}-3 x^{4}\)

Step-by-Step Solution

Answer

F(x) is increasing on (-√2, 0) and (√2, ∞); decreasing on (-∞, -√2) and (0, √2). Concave up on (-∞, -√(6/5)) and (√(6/5), ∞); concave down on (-√(6/5), 0) and (0, √(6/5)).

Step 1: Find the First Derivative

First, let's find the first derivative of the function \(F(x) = x^6 - 3x^4\). The first derivative \(F'(x)\) will help us determine where the function is increasing or decreasing. Differentiate each term:\[F'(x) = \frac{d}{dx}(x^6) - \frac{d}{dx}(3x^4) = 6x^5 - 12x^3\]

Step 2: Solve for Critical Points

Set the first derivative equal to zero to find critical points, which will help identify intervals of increase and decrease:\[6x^5 - 12x^3 = 0\]Factor out the common term:\[6x^3(x^2 - 2) = 0\]This gives critical points at \(x = 0\) and \(x = \pm \sqrt{2}\).

Step 3: Test Intervals for Increasing/Decreasing

Use the critical points to divide the number line into intervals: \((-\infty, -\sqrt{2})\), \((-\sqrt{2}, 0)\), \((0, \sqrt{2})\), and \((\sqrt{2}, \infty)\). Test each interval to see if the function is increasing or decreasing:- Pick \(x = -2\): \(F'(-2) = 6(-2)^5 - 12(-2)^3 = -192\) (decreasing)- Pick \(x = -1\): \(F'(-1) = 6(-1)^5 - 12(-1)^3 = 6\) (increasing)- Pick \(x = 1\): \(F'(1) = 6(1)^5 - 12(1)^3 = -6\) (decreasing)- Pick \(x = 2\): \(F'(2) = 6(2)^5 - 12(2)^3 = 192\) (increasing)

Step 4: Find the Second Derivative

Next, find the second derivative \(F''(x)\) to determine concavity:\[F''(x) = \frac{d}{dx}(6x^5 - 12x^3) = 30x^4 - 36x^2\]Factor to simplify:\[F''(x) = 6x^2(5x^2 - 6)\]

Step 5: Solve for Inflection Points

Set the second derivative to zero to identify points of concavity changes:\[6x^2(5x^2 - 6) = 0\]This gives solutions at \(x = 0\) and \(x = \pm \sqrt{\frac{6}{5}}\).

Step 6: Test Intervals for Concavity

Use the inflection points to test intervals for concavity: \((-\infty, -\sqrt{\frac{6}{5}})\), \((-\sqrt{\frac{6}{5}}, 0)\), \((0, \sqrt{\frac{6}{5}})\), and \((\sqrt{\frac{6}{5}}, \infty)\).- Pick \(x = -2\): \(F''(-2) = 6(-2)^2(5(-2)^2 - 6) = 168\) (concave up)- Pick \(x = -1\): \(F''(-1) = 6(-1)^2(5(-1)^2 - 6) = -6\) (concave down)- Pick \(x = 1\): \(F''(1) = 6(1)^2(5(1)^2 - 6) = -6\) (concave down)- Pick \(x = 2\): \(F''(2) = 6(2)^2(5(2)^2 - 6) = 168\) (concave up)

Step 7: Sketch the Graph

From the intervals, we have:- Increasing on \((-\sqrt{2}, 0)\) and \((\sqrt{2}, \infty)\).- Decreasing on \((-\infty, -\sqrt{2})\) and \((0, \sqrt{2})\).- Concave up on \((-\infty, -\sqrt{\frac{6}{5}})\) and \((\sqrt{\frac{6}{5}}, \infty)\).- Concave down on \((-\sqrt{\frac{6}{5}}, 0)\) and \((0, \sqrt{\frac{6}{5}})\).Using this information, sketch the graph showing these characteristics.

Key Concepts

First Derivative

The first derivative of a function is a powerful tool that helps us determine whether a function is increasing or decreasing on certain intervals. For a given function like \(F(x) = x^6 - 3x^4\), we begin by finding its first derivative, which is denoted as \(F'(x)\).
To do this, we differentiate each term separately: \(F'(x) = 6x^5 - 12x^3\).
This derivative reveals how the function behaves with respect to changes in \(x\).

• If \(F'(x) > 0\) for an interval, the function is increasing on that interval.
• If \(F'(x) < 0\) for an interval, the function is decreasing on that interval.

The first derivative is our primary tool to identify which parts of the function's graph are climbing upwards or sloping downwards.

Critical Points

Critical points are values of \(x\) where the first derivative is either zero or undefined. These points are crucial because they indicate potential "turning points" where a graph changes from increasing to decreasing, or vice versa.
To find them, we set the first derivative equal to zero. For our example function, we solve \(6x^5 - 12x^3 = 0\).
Factoring gives us: \(6x^3(x^2 - 2) = 0\), leading to critical points at \(x = 0\) and \(x = \pm \sqrt{2}\).

• Check around these points to see how the function behaves on either side—this will indicate if they're local minimums or maximums.
• Use these points to divide the number line into segments where we can test the intervals for increasing or decreasing behavior.

Critical points are integral in understanding the graph's flow and finding places where the direction changes.

Concavity

Concavity refers to the way a graph curves, either upwards or downwards. Understanding concavity provides insights into the "bend" of the graph, whether it dips or rises between intervals. To determine this, we examine the second derivative, \(F''(x)\), of our function.
For \(F(x)\), we calculate \(F''(x) = 30x^4 - 36x^2\).

• If \(F''(x) > 0\) on an interval, the graph is concave up like a "smiley face."
• If \(F''(x) < 0\), it is concave down like a "frown."

Concavity helps us predict where a curve might bend further up or down and assists in sketching the graph accurately.

Second Derivative

The second derivative provides another layer of information about how the graph of a function behaves, specifically focusing on its concavity and any changes in that concavity. To find significant points, set \(F''(x) = 0\). In our function's case, solving \(30x^4 - 36x^2 = 0\) gives inflection points where potential changes in concavity occur.
Via factoring, we find \(6x^2(5x^2 - 6) = 0\), resulting in solutions at \(x = 0\) and \(x = \pm \sqrt{\frac{6}{5}}\).

• The inflection points divide the graph into segments which we test for concavity.
• An understanding of second derivatives enhances our ability to visualize the "twist" and "turn" of the function.

The analysis of the second derivative provides detailed information about the "shape" and structure of the graph, allowing you to complete a more accurate sketch.