Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-18, use the Concavity Theorem to determine where the given function is concave up and where it is concave down. Also find all inflection points. \(f(x)=x^{4}+8 x^{3}-2\)

Step-by-Step Solution

Answer

Concave up: \((-\infty, -4) \cup (0, \infty)\), concave down: \((-4, 0)\); Inflection points: \((-4, -258), (0, -2)\).

Step 1: Find the First Derivative

First, we need to find the first derivative of the function to proceed with finding the second derivative later on. The function is given as \( f(x) = x^4 + 8x^3 - 2 \). Use the power rule to differentiate:\[f'(x) = 4x^3 + 24x^2\]

Step 2: Find the Second Derivative

Next, find the second derivative of \( f(x) \), which will help in determining concavity. Differentiate \( f'(x) = 4x^3 + 24x^2 \):\[f''(x) = 12x^2 + 48x\]

Step 3: Solve for Critical Points of the Second Derivative

To find the critical points that might be inflection points, set the second derivative equal to zero and solve for \( x \):\[12x^2 + 48x = 0\]Factor the equation:\[12x(x + 4) = 0\]Thus, \( x = 0 \) or \( x = -4 \).

Step 4: Analyze the Sign of the Second Derivative

Determine where \( f''(x) > 0 \) or \( f''(x) < 0 \) by testing intervals defined by \( x = 0 \) and \( x = -4 \):- Choose \( x = -5 \) for \( (-\infty, -4) \): \( f''(-5) = 12(-5)^2 + 48(-5) = 300 - 240 = 60 > 0 \)- Choose \( x = -2 \) for \( (-4, 0) \): \( f''(-2) = 12(-2)^2 + 48(-2) = 48 - 96 = -48 < 0 \)- Choose \( x = 1 \) for \( (0, \infty) \): \( f''(1) = 12(1)^2 + 48(1) = 60 > 0 \)

Step 5: Determine Intervals of Concavity

Based on the sign analysis of the second derivative:- \( f(x) \) is concave up on \( (-\infty, -4) \cup (0, \infty) \)- \( f(x) \) is concave down on \( (-4, 0) \)

Step 6: Identify Inflection Points

An inflection point occurs where the concavity changes. From Step 4, these changes occur at \( x = -4 \) and \( x = 0 \). Evaluate \( f(x) \) at these \( x \) values:- \( f(-4) = (-4)^4 + 8(-4)^3 - 2 = 256 - 512 - 2 = -258 \)- \( f(0) = 0^4 + 8(0)^3 - 2 = -2 \)Thus, the points \((-4, -258)\) and \((0, -2)\) are inflection points.

Key Concepts

Inflection Points

Inflection points are special points on the graph of a function where the concavity changes. The change in concavity means that the graph switches from being concave up to concave down, or vice versa.
To find inflection points, we look for places where the second derivative changes sign. This is because the second derivative tells us about the concavity of the function. We solve for values of the variable by setting the second derivative equal to zero or undefined and finding those specific points.
In our example, the second derivative is given as:

• \(f''(x) = 12x^2 + 48x\)

Setting this equal to zero helps us find possible inflection points:

• \(12x(x + 4) = 0\)
• This gives us the solutions \(x = 0\) and \(x = -4\).

These solutions tell us that the graph may change concavity at \(x = 0\) and \(x = -4\). By evaluating the original function at these points, we find the inflection points to be \((-4, -258)\) and \((0, -2)\).

Concave Up

A function is said to be concave up on an interval if its graph bends upwards, like the shape of a cup. We use the second derivative to determine where this occurs because a positive second derivative indicates concave up.
Consider our function again, with the second derivative:

• \(f''(x) = 12x^2 + 48x\)

For the intervals specified by the values \(x = -4\) and \(x = 0\), we test and analyze using test points. For concavity, if \(f''(x) > 0\), the graph is concave up:

• In our example, test with \(x = -5\) yields \(f''(-5) = 60 > 0\), meaning concave up on \((-\infty, -4)\).
• Similarly, test with \(x = 1\) yields \(f''(1) = 60 > 0\), indicating concave up on \((0, \infty)\).

So, the function is concave up on the intervals \((-\infty, -4)\) and \((0, \infty)\).

Concave Down

A function is concave down on an interval when its graph appears to bend downwards, resembling an upside down cup. The second derivative helps identify these intervals, as a function is concave down where its second derivative is negative.
To identify concave down regions, look at \(f''(x) = 12x^2 + 48x\).
From the previous analysis, select intervals to test:

• Between \(x = -4\) and \(x = 0\), use \(x = -2\). Here, \(f''(-2) = -48 < 0\), meaning the function is concave down.

Thus, the function \(f(x)\) is concave down on the interval \((-4, 0)\). This tells us there’s a change in the bending direction of the curve between these points, further confirming potential inflection points.

Second Derivative

The second derivative of a function is a derivative of the derivative. It provides insight into the concavity and the rate of change of the slope of the function's graph.
For the function \(f(x) = x^4 + 8x^3 - 2\), the second derivative is:

• \(f''(x) = 12x^2 + 48x\)

This is crucial in determining where the function is concave up or down. By analyzing the signs of the second derivative across different intervals, we can make conclusions about the graph’s shape.

• If \(f''(x) > 0\), the graph is concave up.
• If \(f''(x) < 0\), the graph is concave down.

Additionally, finding the x-values where \(f''(x) = 0\) or changes sign is key to locating inflection points. This is why the second derivative holds great importance when graphing and analyzing functions.