Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

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

The function is concave up on \((- fty, -4) \cup (0, fty)\) and concave down on \((-4, 0)\). Inflection points are \((-4, -258)\) and \((0, -2)\).

Step 1: Find the First Derivative

To find where the function is concave up or down, we first need to find its derivatives. The first derivative of the function \( f(x) = x^4 + 8x^3 - 2 \) is found by applying the power rule: \[ f'(x) = 4x^3 + 24x^2. \]

Step 2: Find the Second Derivative

Next, we must calculate the second derivative to apply the Concavity Theorem. Differentiating the first derivative \( f'(x) = 4x^3 + 24x^2 \), we get: \[ f''(x) = 12x^2 + 48x. \]

Step 3: Find Critical Points of the Second Derivative

To use the Concavity Theorem, we need to find where the second derivative is zero or undefined (critical points). Set the second derivative equal to zero and solve: \[ 12x^2 + 48x = 0. \] Factor out the common term: \[ 12x(x + 4) = 0. \] This gives the solutions \( x = 0 \) and \( x = -4 \).

Step 4: Determine Intervals for Concavity

The critical points divide the real number line into intervals. Check the sign of \( f''(x) \) in each interval to determine concavity:- For \( x < -4 \), choose \( x = -5 \): \( f''(-5) = 12(-5)^2 + 48(-5) > 0 \). Thus, \( f(x) \) is concave up.- For \( -4 < x < 0 \), choose \( x = -1 \): \( f''(-1) = 12(-1)^2 + 48(-1) = -36 < 0 \). Thus, \( f(x) \) is concave down.- For \( x > 0 \), choose \( x = 1 \): \( f''(1) = 12(1)^2 + 48(1) > 0 \). Thus, \( f(x) \) is concave up.

Step 5: Identify Inflection Points

Inflection points occur where the concavity changes, which happens at the critical points \( x = -4 \) and \( x = 0 \). To find the coordinates of the inflection points, substitute back into the original function:- At \( x = -4 \): \( f(-4) = (-4)^4 + 8(-4)^3 - 2 = 256 - 512 - 2 = -258 \), giving the point \((-4, -258)\).- At \( x = 0 \): \( f(0) = 0^4 + 8(0)^3 - 2 = -2 \), giving the point \((0, -2)\).

Key Concepts

First Derivative

The first derivative is crucial for understanding how a function behaves. It tells us the rate at which the function is changing at any point. For the function \[ f(x) = x^4 + 8x^3 - 2, \]the first derivative is obtained using the power rule: \[ f'(x) = 4x^3 + 24x^2. \] This derivative helps us see where the original function is increasing or decreasing.

• If the first derivative is positive, the function is increasing at that point.
• If it's negative, the function is decreasing.
• Where it's zero, we potentially have a maximum or minimum point, subject to further analysis with the second derivative.

Understanding the first derivative helps set the stage for deeper analysis using the second derivative needed for the concavity test.

Second Derivative

The second derivative tells us about the function's concavity — whether it bows upwards or downwards. Continuing from our first derivative,\[ f'(x) = 4x^3 + 24x^2, \]the second derivative is calculated as:\[ f''(x) = 12x^2 + 48x. \]When analyzing this new derivative:

• If the second derivative is positive over an interval, the function is concave up like a smile.
• If it is negative, the function is concave down like a frown.

Applying these concepts can help us in identifying potential inflection points based on changes in concavity.

Critical Points

Critical points of the second derivative are key to understanding where changes in concavity might occur. These points come from solving \[ f''(x) = 0 \]or where the second derivative does not exist. For\[ f''(x) = 12x^2 + 48x, \]we set it equal to zero, resulting in:\[ 12x(x + 4) = 0. \]The solutions are the critical points:

• \( x = 0 \)
• \( x = -4 \)

These points split the number line into intervals where we can check concavity and locate inflection points. Critical points of the first derivative indicate potential maxima and minima, while those of the second derivative hint at inflection points.

Inflection Points

Inflection points are where a function changes its concavity. They occur at the critical points where the sign of the second derivative changes. Calculations have shown two critical points:

• \( x = -4 \)
• \( x = 0 \)

Let's evaluate these for inflection:

• At \( x = -4 \), plug into the original function: \[ f(-4) = (-4)^4 + 8(-4)^3 - 2 = -258, \]thus, the inflection point is \((-4, -258)\).
• At \( x = 0 \), we get: \[ f(0) = 0^4 + 8(0)^3 - 2 = -2, \]providing \((0, -2)\) as another inflection point.

Finding these points on a graph reveals where the curve flexes, switching from a concave up to a concave down shape or vice versa, showing a critical behavior change in the function's geometry.