Problem 11 - Calculus Single Variable

Question

Determine the intervals on which the given function \(f\) is concave up, the intervals on which \(f\) is concave down, and the points of inflection of \(f\). Find all critical points. Use the Second Derivative Test to identify the points \(x\) at which \(f(x)\) is a local minimum value and the points at which \(f(x)\) is a local maximum value. $$ f(x)=x^{4}-2 x^{3}-9 x^{2} $$

Step-by-Step Solution

Answer

Concave up: \( (-\infty, \frac{1-\sqrt{7}}{2}) \cup (\frac{1+\sqrt{7}}{2}, \infty) \); Concave down: \( (\frac{1-\sqrt{7}}{2}, \frac{1+\sqrt{7}}{2}) \); Inflection points at \( x = \frac{1 \pm \sqrt{7}}{2} \); Critical points at \( x = -1.5, 0, 3 \). Local max at \( x=0 \) by second derivative test.

Step 1: Find the First Derivative

To analyze the function's concavity and critical points, we first find the first derivative of the function. Given, \[ f(x) = x^4 - 2x^3 - 9x^2 \]The first derivative is:\[ f'(x) = \frac{d}{dx}(x^4) - \frac{d}{dx}(2x^3) - \frac{d}{dx}(9x^2) = 4x^3 - 6x^2 - 18x \]

Step 2: Find the Second Derivative

Next, we find the second derivative to determine intervals of concavity and potential inflection points. Differentiate \( f'(x) \):\[ f''(x) = \frac{d}{dx}(4x^3 - 6x^2 - 18x) = 12x^2 - 12x - 18 \]

Step 3: Determine Concavity Intervals

Set the second derivative equal to zero to find potential inflection points:\[ 12x^2 - 12x - 18 = 0 \]Simplifying, we have:\[ x^2 - x - \frac{3}{2} = 0 \]Using the quadratic formula, \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a=1, b=-1, c=-\frac{3}{2} \), we get:\[ x = \frac{1 \pm \sqrt{1 + 6}}{2} = \frac{1 \pm \sqrt{7}}{2} \]These are the potential inflection points.

Step 4: Test Intervals for Concavity

Select test points in the intervals determined by the points \( x = \frac{1 \pm \sqrt{7}}{2} \) to check where \( f''(x) > 0 \) (concave up) and \( f''(x) < 0 \) (concave down). For example, use points like \( x = -1, 0, 2 \):- For \( x < \frac{1 - \sqrt{7}}{2} \), choose \( x = -1 \): \( f''(-1) = 42 > 0 \), concave up.- For \( \frac{1 - \sqrt{7}}{2} < x < \frac{1 + \sqrt{7}}{2} \), choose \( x = 0 \): \( f''(0) = -18 < 0 \), concave down.- For \( x > \frac{1 + \sqrt{7}}{2} \), choose \( x = 2 \): \( f''(2) = 6 > 0 \), concave up.

Step 5: Find Critical Points

Set the first derivative equal to zero to find critical points:\[ 4x^3 - 6x^2 - 18x = 0 \]Factor out common terms:\[ 2x(2x^2 - 3x - 9) = 0 \]We solve:1. \( 2x = 0 \) gives \( x = 0 \).2. \( 2x^2 - 3x - 9 = 0 \) using the quadratic formula gives roots \( x = \frac{3 \pm \sqrt{81}}{4} \), getting \( x = 3 \) or \( x = -1.5 \).

Step 6: Use Second Derivative Test for Extrema

Apply the second derivative test:- For \( x=0 \): \( f''(0) = -18 \), \( f''(0) < 0 \), indicating a local maximum.- For \( x=-1.5 \): Find \( f''(-1.5) \) (positive or negative determines max/min).- For \( x=3 \): Find \( f''(3) \) (positive or negative determines max/min).

Key Concepts

Concavity

Concavity helps us understand the shape of the graph of a function. To determine concavity, we look at the sign of the second derivative, \( f''(x) \). If \( f''(x) > 0 \), the curve is concave up, resembling a cup, and if \( f''(x) < 0 \), the curve is concave down, resembling a frown. Concavity tells us whether a segment of the graph is curving upwards or downwards.
To find intervals of concavity, find the second derivative and solve for when it equals zero to find potential points of inflection. You'll then test intervals between these points to see where \( f''(x) \) is positive or negative.

• Concave up: \( f''(x) > 0 \)
• Concave down: \( f''(x) < 0 \)

This method reveals sections of the graph that are bending in different directions, giving us insights into its rising and falling behavior.

Critical Points

A critical point occurs where the first derivative, \( f'(x) \), is either zero or undefined. Critical points are crucial because they can indicate where a function might reach a maximum or minimum value.
To find critical points, solve \( f'(x) = 0 \). Factor the resulting expression to identify the points. These points are where the slope of the tangent to the curve is zero, meaning the curve could be shifting from increasing to decreasing or vice versa. In our example, critical points were found by solving:

• \( 4x^3 - 6x^2 - 18x = 0 \)
• \( 2x(2x^2 - 3x - 9) = 0 \)

Identifying critical points allows us to locate potential maximums and minimums, which are essential in analyzing the overall behavior of the function.

Inflection Points

Inflection points occur where the function changes concavity—from concave up to concave down or vice versa. They are found where the second derivative changes its sign, and are identified by setting \( f''(x) = 0 \).
In our example, after finding \( f''(x) \), we set:
\[ 12x^2 - 12x - 18 = 0 \]
This led to potential inflection points by solving it, often using the quadratic formula. These points are critical for understanding the dynamic nature of the graph as it changes shape.

• Change in concavity
• Found at roots of \( f''(x) \)

Understanding inflection points helps us grasp how and where the function's graph curves.

Local Maximum and Minimum

Local maximum and minimum points are where a function reaches its highest or lowest value within a certain interval. These can be identified using the Second Derivative Test at the critical points found.
To apply the Second Derivative Test:

• If \( f''(x) > 0 \) at a critical point, then \( f(x) \) has a local minimum there.
• If \( f''(x) < 0 \) at a critical point, then \( f(x) \) has a local maximum there.

In our solution, by inserting each critical point into the second derivative, we determined whether each is a local peak or valley. Recognizing these points is crucial for sketching and understanding the function's behavior.

First Derivative

The first derivative, \( f'(x) \), represents the slope of the tangent line to the curve at any given point. It tells us how the function's value is changing immediately at that point—whether it's increasing or decreasing.
To find \( f'(x) \), differentiate the original function, an essential step in locating critical points. In our example:

• Differentiate: \( f'(x) = 4x^3 - 6x^2 - 18x \)

This derivative assists in determining intervals where the function is either rising or falling, and aids in identifying critical points for further analysis of potential peaks and troughs.

Quadratic Formula

The quadratic formula is a tool to solve quadratic equations of the form \( ax^2 + bx + c = 0 \). It provides solutions to find roots when simple factoring isn't possible or easy.
The formula is:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

In our scenario, it was used to solve for the roots of the quadratic expression derived from setting the second derivative to zero, \( 12x^2 - 12x - 18 = 0 \). Applying the quadratic formula when a quadratic isn't factorable provides a straightforward method to find those tricky roots, crucial for further analysis like identifying inflection points in this context.