Problem 232 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing, b. local minima and maxima of \(f\), c. intervals where \(f\) is concave up and concave down, and d. the inflection points of \(f\). Sketch the curve, then use a calculator to compare your answer. If you cannot determine the exact answer analytically, use a calculator. \(f(x)=x+\sin (2 x)\) over \(x=\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]\)

Step-by-Step Solution

Answer

The function is always increasing; no local extrema; concave up on \((-\pi/2, 0)\), concave down on \((0, \pi/2)\); inflection point at \( (0, 0) \).

Step 1: Find the First Derivative

To determine intervals where the function is increasing or decreasing, we first need to find its first derivative. The function is given as \( f(x) = x + \sin(2x) \). Using derivative rules, we find:\[ f'(x) = 1 + 2\cos(2x) \]

Step 2: Determine Critical Points

To find the critical points, set the first derivative equal to zero and solve for \( x \):\[ 1 + 2\cos(2x) = 0 \]Solving for \( x \), we get:\[ \cos(2x) = -\frac{1}{2} \]The solutions to this equation within the interval \((-\pi/2, \pi/2)\) are \( x = -\pi/3 \) and \( x = \pi/3 \). These are the critical points.

Step 3: Determine Intervals of Increase and Decrease

We will test intervals obtained from the critical points to check if the function is increasing or decreasing:- Between \( x = -\pi/2 \) and \( x = -\pi/3 \), test a point like \( x = -\pi/4 \): \[ f'(-\pi/4) = 1 + 2\cos(-\pi/2) = 1 \] (positive, hence increasing)- Between \( x = -\pi/3 \) and \( x = \pi/3 \), test a point like \( x = 0 \): \[ f'(0) = 1 + 2\cos(0) = 3 \] (positive, hence increasing)- Between \( x = \pi/3 \) and \( x = \pi/2 \), test a point like \( x = \pi/4 \): \[ f'(\pi/4) = 1 + 2\cos(\pi/2) = 1 \] (positive, hence increasing)Thus, the function is increasing on the entire interval \([-\pi/2, \pi/2]\).

Step 4: Identify Local Minima and Maxima

Since the function is increasing throughout the interval, there are no local minima or maxima on \([-\pi/2, \pi/2]\).

Step 5: Find the Second Derivative

Now, we find the second derivative to determine concavity:\[ f''(x) = -4\sin(2x) \]

Step 6: Determine Concavity Intervals

Evaluate the sign changes of \( f''(x) = -4\sin(2x) \) in the given interval:- Find points where \( f''(x) = 0 \): \[ \sin(2x) = 0 \] The solutions in the interval \([-\pi/2, \pi/2]\) are \( x = 0 \). These are potential inflection points.- Test intervals identified by solving \( \sin(2x) = 0 \): - Between \( x = -\pi/2 \) and \( x = 0 \), test \( x = -\pi/4 \): \[ f''(-\pi/4) = -4\sin(-\pi/2) = 4 \] (positive, concave up) - Between \( x = 0 \) and \( x = \pi/2 \), test \( x = \pi/4 \): \[ f''(\pi/4) = -4\sin(\pi/2) = -4 \] (negative, concave down)The function is concave up on \((-\pi/2, 0)\) and concave down on \((0, \pi/2)\).

Step 7: Identify Inflection Points

Since \( f''(x) = 0 \) at \( x = 0 \), and the sign of \( f''(x) \) changes around \( x = 0 \), \( x = 0 \) is an inflection point. The corresponding \( f(x) \) value:\[ f(0) = 0 + \sin(0) = 0 \]So, the inflection point is \( (0, 0) \).

Key Concepts

First Derivative

When studying calculus, the first derivative of a function is crucial for understanding how that function behaves. For a function like \(f(x) = x + \sin(2x)\), the first derivative \(f'(x)\) describes how \(f(x)\) changes with \(x\). To find \(f'(x)\), we differentiate each term: the derivative of \(x\) is simply 1, and using the chain rule, the derivative of \(\sin(2x)\) is \(2\cos(2x)\). Thus, the first derivative is:

• \(f'(x) = 1 + 2\cos(2x)\)

The first derivative tells us about the slope or rate of change of the original function. By analyzing this derivative, we can begin to understand where the function is increasing or decreasing.
It's often helpful to visualize the behavior of the function by knowing the slope at various points along its domain.

Critical Points

Critical points occur where a function's first derivative is zero or undefined. These points are essential because they might indicate a function's local maximum or minimum values. For \(f(x) = x + \sin(2x)\), we determine the critical points by setting the first derivative \(f'(x) = 1 + 2\cos(2x)\) to zero:

• \(1 + 2\cos(2x) = 0\)

Solving for \(x\), we'll find that \( \cos(2x) = -\frac{1}{2}\). Within the interval \([-\pi/2, \pi/2]\), the solutions for this equation are \( x = -\pi/3 \) and \( x = \pi/3 \).
These values split the interval into segments we can examine to understand the behavior of the function around these critical points, such as changes from increasing to decreasing.

Second Derivative

The second derivative \(f''(x)\) of a function gives us information about the concavity of the graph of \(f\). For \(f(x) = x + \sin(2x)\), we found the first derivative was \(f'(x) = 1 + 2\cos(2x)\). Differentiating again gives us the second derivative:

• \(f''(x) = -4\sin(2x)\)

The second derivative helps identify how the rate of change of \(f(x)\) itself is changing—whether it is accelerating upwards or downwards. By testing the sign of \(f''(x)\) at various points, we gain insight into where the graph is concave up or concave down.
This is crucial for finding points of inflection, where the concavity changes.

Concavity

Understanding a function's concavity can give significant insights into the shape of its graph. When a graph is concave up, it bends upwards like a cup; if it's concave down, it bends downwards like a frown. For \(f(x) = x + \sin(2x)\), we use the second derivative \(f''(x) = -4\sin(2x)\) to determine concavity.

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

By evaluating \(f''(x)\) within the interval \([-\pi/2, \pi/2]\), we found that the function is:

• Concave up on \((-\pi/2, 0)\)
• Concave down on \((0, \pi/2)\)

Thus, the concavity gives us an understanding of the overall behavior and shape of the function's graph, showing where it turns or curves in each direction.

Inflection Point

An inflection point is a point on the curve where the concavity changes sign. For \(f(x) = x + \sin(2x)\), we determined the inflection points by finding where the second derivative \(f''(x) = -4\sin(2x)\) is zero or changes sign.

• \(f''(x) = 0\) gives us potential inflection points.

In this case, the equation \(\sin(2x) = 0\) provided the solution \(x = 0\) within the interval \([-\pi/2, \pi/2]\).At \(x = 0\), the surrounding signs of \(f''(x)\) change from positive to negative, confirming that \(x = 0\) is indeed an inflection point.
The coordinates of this point are \((0, 0)\), where the graph of the function transitions from concave up to concave down, highlighting a significant turning behavior in the graph's shape.