Problem 17 - Calculus - AP Edition

Question

Find: (a) the intervals on which \(f\) is increasing, (b) the intervals on which \(f\) is decreasing, (c) the open intervals on which \(f\) is concave up, (d) the open intervals on which \(f\) is concave down, and (e) the \(x\) -coordinates of all inflection points. $$f(x)=(2 x+1)^{3}$$

Step-by-Step Solution

Answer

(a) Increasing on \((- \infty, \infty)\); (b) None; (c) \((-\frac{1}{2}, \infty)\); (d) \((- \infty, -\frac{1}{2})\); (e) \(x = -\frac{1}{2}\).

Step 1: Compute the First Derivative

To find where the function is increasing or decreasing, first compute the derivative of \(f(x)\). Let \(f(x) = (2x + 1)^3\). Using the chain rule, we find the derivative:\[ f'(x) = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2 \].

Step 2: Find Critical Points

Set the first derivative equal to zero and solve for \(x\) to find the critical points:\[ 6(2x + 1)^2 = 0 \]. Since this equation has no solution (as \((2x + 1)^2\) is never zero), there are no critical points where \(f(x)\) changes from increasing to decreasing or vice versa.

Step 3: Determine Intervals of Increase/Decrease

Since there are no critical points, analyze the sign of \(f'(x) = 6(2x + 1)^2\). This expression is always positive for all real numbers, thus \(f\) is increasing on \((-\infty, \infty)\) and there are no intervals where \(f\) is decreasing.

Step 4: Compute the Second Derivative

Now, find the second derivative to analyze concavity. Start by differentiating the first derivative:\[ f'(x) = 6(2x + 1)^2 \].Apply the chain rule:\[ f''(x) = 12(2x + 1) \cdot 2 = 24(2x + 1) \].

Step 5: Analyze Concavity

Determine the intervals where \(f''(x)\) is positive or negative:- \(f''(x) > 0\) when \(2x + 1 > 0\), or \(x > -\frac{1}{2}\), which is concave up.- \(f''(x) < 0\) when \(2x + 1 < 0\), or \(x < -\frac{1}{2}\), which is concave down.

Step 6: Find Inflection Points

An inflection point is where the concavity changes. This occurs when \(f''(x)\) changes sign:- Solve \(f''(x) = 0\):\[ 24(2x + 1) = 0 \]\[ x = -\frac{1}{2} \].Therefore, the \(x\)-coordinate of the inflection point is \(-\frac{1}{2}\).

Key Concepts

First Derivative

The first derivative of a function, often denoted as \( f'(x) \), tells us about the rate of change of the function. Specifically, it helps to identify intervals where the function is increasing or decreasing.
To find the first derivative of \( f(x) = (2x + 1)^3 \), apply the chain rule. The calculation would look like this:

• Differentiate the outer function: \( u^3 \) becomes \( 3u^2 \).
• Multiply by the derivative of the inner function \( u = 2x + 1 \): this gives \( 2 \).

Thus, the computation yields \( f'(x) = 6(2x + 1)^2 \).
This derivative is positive for all real \( x \), meaning the function is always increasing.
There are no critical points from solving \( f'(x) = 0 \) since the square of any real number is never zero.

Second Derivative

The second derivative of a function, denoted \( f''(x) \), provides information about the concavity of the function. Concavity refers to the direction in which the function curves.
We find the second derivative by differentiating the first derivative. Using the function \( f'(x) = 6(2x + 1)^2 \), we apply the chain rule again:

• Differentiate \( 6(2x + 1)^2 \) using \( 12(2x + 1) \).
• Multiply by the derivative of \( 2x + 1 \), which is \( 2 \).

This gives us \( f''(x) = 24(2x + 1) \).
This expression helps us understand where the function is concave up or down, leading us into the concavity analysis.

Concavity Analysis

Concavity analysis involves determining where a function is concave up or concave down based on the sign of its second derivative.
Here's a simple way to break it down:

• If \( f''(x) > 0 \), the function is concave up. It resembles a U-shape with a smile.
• If \( f''(x) < 0 \), the function is concave down. It resembles an upside-down U or a frown.

For our function \( f''(x) = 24(2x + 1) \):

• \( f''(x) > 0 \) when \( x > -\frac{1}{2} \); hence, the function is concave up on \( (-\frac{1}{2}, \infty) \).
• \( f''(x) < 0 \) when \( x < -\frac{1}{2} \); thus, the function is concave down on \( (-\infty, -\frac{1}{2}) \).

Inflection Points

An inflection point is where the function changes concavity, from concave up to concave down or vice versa.
To find inflection points, locate where the second derivative \( f''(x) \) changes sign. This occurs when \( f''(x) = 0 \).

• Solve \( 24(2x + 1) = 0 \).
• Find \( x = -\frac{1}{2} \).

At \( x = -\frac{1}{2} \), the second derivative shifts from negative to positive. Therefore, this is the inflection point. It marks the transition between concave down and concave up, illustrating a significant change in the function's curvature.