Problem 221 - Calculus Volume 1

Question

Determine a. intervals where \(f\) is increasing or decreasing and b. local minima and maxima of \(f\). \(f(x)=\sin x+\sin ^{3} x\) over \(-\pi

Step-by-Step Solution

Answer

Increasing on \((-\frac{\pi}{2}, \frac{\pi}{2})\), decreasing on \((-\pi, -\frac{\pi}{2})\) and \((\frac{\pi}{2}, \pi)\). Local minimum at \(x = -\frac{\pi}{2}\), local maximum at \(x = \frac{\pi}{2}\).

Step 1: Find the Derivative

To determine where a function is increasing or decreasing, we first need the derivative of the function. Calculate the derivative of \(f(x) = \sin x + \sin^3 x\). Use the chain rule: \[ f'(x) = \frac{d}{dx} [\sin x] + \frac{d}{dx} [\sin^3 x] = \cos x + 3\sin^2 x \cdot \cos x. \]Factor out \(\cos x\) from the derivative:\[ f'(x) = \cos x (1 + 3\sin^2 x). \]

Step 2: Critical Points and Testing Intervals

Set \(f'(x) = 0\) to find critical points:\[ \cos x (1 + 3\sin^2 x) = 0. \]This yields two conditions: \(\cos x = 0\) or \(1 + 3\sin^2 x = 0\). However, \(1 + 3\sin^2 x = 0\) has no real solutions as \(\sin^2 x\) is always non-negative. Thus, we solve \(\cos x = 0\) to find \(x = -\frac{\pi}{2}\) and \(x = \frac{\pi}{2}\).These points divide the interval \((-\pi, \pi)\) into three intervals: \((-\pi, -\frac{\pi}{2})\), \((-\frac{\pi}{2}, \frac{\pi}{2})\), and \((\frac{\pi}{2}, \pi)\).

Step 3: Test Sign of Derivative

For \((-\pi, -\frac{\pi}{2})\): Choose \(x = -\frac{3\pi}{4}\). Here \(\cos x < 0\), so \(f'(x) < 0\) (function is decreasing).For \((-\frac{\pi}{2}, \frac{\pi}{2})\): Choose \(x = 0\). Here \(\cos x > 0\), so \(f'(x) > 0\) (function is increasing).For \((\frac{\pi}{2}, \pi)\): Choose \(x = \frac{3\pi}{4}\). Here \(\cos x < 0\), so \(f'(x) < 0\) (function is decreasing).

Step 4: Identify Local Extrema

Analyze the behavior at critical points to find local extrema:- At \(x = -\frac{\pi}{2}\), the derivative changes from negative to positive, indicating a local minimum.- At \(x = \frac{\pi}{2}\), the derivative changes from positive to negative, indicating a local maximum.

Key Concepts

Derivative of trigonometric functions

In this exercise, we need to find the derivative of the function \(f(x) = \sin x + \sin^3 x\). Calculating derivatives of trigonometric functions is a fundamental skill in calculus. For basic trigonometric functions like \( \sin x \) and \( \cos x \), their derivatives are well-known.

• The derivative of \( \sin x \) is \( \cos x \).
• The derivative of \( \cos x \) is \(-\sin x \).

Plus, when dealing with a composite function like \( \sin^3 x \), we need to apply the chain rule. In this case, the chain rule helps us handle the power of a sine function. We first find the derivative of the outer function (raising to a power) and multiply it by the derivative of the inner function (\(\sin x\)).
Hence, "\(\frac{d}{dx} [\sin^3 x] = 3\sin^2 x \cdot \cos x\)". Combining these, we derive the entire function as \(f'(x) = \cos x + 3\sin^2 x \cdot \cos x\). Understanding the chain rule and these trigonometric derivatives is crucial for solving calculus problems effectively.

Critical points analysis

Critical points in a function occur where the derivative is zero or undefined. They are pivotal in determining where the function might change its increasing or decreasing behavior. For the function \(f(x) = \sin x + \sin^3 x\), we have derived the derivative as \(f'(x) = \cos x(1 + 3\sin^2 x)\).
To find the critical points, we set the derivative equal to zero:

• \(f'(x) = 0\).

This yields the equation \(\cos x(1 + 3\sin^2 x) = 0\). Further analysis leads us to solve for \(\cos x = 0\), identifying critical points at \(x = -\frac{\pi}{2}\) and \(x = \frac{\pi}{2}\).
Note that the term "\(1 + 3\sin^2 x = 0\)" is not possible because \(\sin^2 x\) is always non-negative. Thus, we focus on these values where \(\cos x\) vanishes. These critical points are essential as they divide the interval \((-\pi, \pi)\) into specific segments for further testing.

Local minima and maxima determination

After finding the critical points, the next step is to determine whether they correspond to local minima or maxima. This involves examining the sign of the derivative in intervals around the critical points.
For \(x = -\frac{\pi}{2}\), the derivative changes from negative to positive as you cross the critical point, indicating a local minimum. Conversely, at \(x = \frac{\pi}{2}\), the derivative switches from positive to negative, signifying a local maximum.

• Local Minimum: \(x = -\frac{\pi}{2}\)
• Local Maximum: \(x = \frac{\pi}{2}\)

By observing these changes, we can confidently identify the nature of these points. This form of analysis—checking the sign of the derivative before and after the critical points—helps us sketch how the function behaves in its domain.

Chain rule in differentiation

The chain rule is an essential differentiation technique, especially when dealing with compounded or nested functions. In our function \(f(x) = \sin x + \sin^3 x\), the \(\sin^3 x\) term requires us to use the chain rule effectively because it contains a function raised to a power, namely \(\sin x\), cubed.
The chain rule states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function. In mathematical terms, if \(y = g(u)\) and \(u = h(x)\), then \(\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}\).
In this problem:

• Outer function "\( (u^3) \)" gives \(3\sin^2 x\).
• Inner function "\( \sin x \)" gives \(\cos x\).

Thus, combining these, we find "\(\frac{d}{dx} [\sin^3 x] = 3\sin^2 x \cdot \cos x\)". Using this rule correctly simplifies differentiation tasks and is indispensable for tackling advanced calculus problems efficiently.