Problem 59 - Single Variable Calculus

Question

Show that the equation \(x+\sin x=b\) has no positive root if \(b<0\) and has one positive root if \(b>0\). Hint: Show that \(f(x)=x+\sin x-b\) is increasing and that \(f(0)>0\) if \(b<0\) and \(f(0)<0\) if \(b>0 .\)

Step-by-Step Solution

Answer

The function \(f(x) = x + \sin x - b\) is increasing as its derivative \(f'(x) = 1 + \cos x \geq 0\). For \(b < 0\), \(f(0) > 0\) and since the function is increasing, there can be no positive roots. When \(b > 0\), \(f(0) < 0\) and as the function is increasing, there will be one positive root.

Step 1: Understanding the given function

We are given the function \(f(x) = x + \sin x - b\), and have to show that there is no positive root for this function if \(b < 0\), and that there is one positive root if \(b > 0\). Let's find the derivative of the given function, which will help us determine if it's increasing or decreasing.

Step 2: Determining whether the function is increasing

The derivative of the function will help us analyze if the function is increasing or decreasing: The derivative of the function is given by \(f'(x) = \frac{df(x)}{dx} = \frac{d(x + \sin x - b)}{dx}\). To find the derivative of \(f(x)\), we’ll use the sum rule for derivatives and the basic rules of differentiation: \(f'(x) = \frac{dx}{dx} + \frac{d(\sin x)}{dx} - \frac{db}{dx}\). Since \(x\) and \(b\) are constants, their derivatives with respect to \(x\) will be 1 and 0, respectively. The derivative of \(\sin x\) is \(\cos x\): \(f'(x) = 1 + \cos x\). We know that the cosine function has a range of \(-1\) to \(1\). Therefore, the minimum value of \(\cos x\) is \(-1\). \(f'(x) = 1 + \cos x \geq 1 - 1 = 0\). Since \(f'(x) \geq 0\), the function \(f(x)\) is increasing.

Step 3: Analyze the function for different values of \(b\)

Now, we need to analyze the behaviour of the function based on the value of \(b\). 1. When \(b < 0\): We will check the value of the function at \(x = 0\): \(f(0) = 0 + \sin0 - b\) Since \(\sin0 = 0\), the above equation becomes: \(f(0) = -b\) Since \(b < 0\), \(f(0) = -b > 0\). As the function is increasing, there cannot be any positive roots for the given equation. 2. When \(b > 0\): We will again check the value of the function at \(x = 0\): \(f(0) = 0 + \sin0 - b\) Since \(\sin0 = 0\), the above equation becomes: \(f(0) = -b\) Since \(b > 0\), \(f(0) = -b < 0\). As the function is increasing, there will be one positive root for the given equation.

Step 4: Conclusion

We have shown that the function \(f(x) = x + \sin x - b\) is increasing. We have also found that if \(b < 0\), \(f(0) > 0\) and there can be no positive roots, and if \(b > 0\), \(f(0) < 0\) and there will be one positive root.

Key Concepts

Derivative of a Function

The derivative of a function represents the rate at which the function's value changes with respect to its variable. It's a fundamental tool in calculus that we use to understand the behavior of functions, including their increasing or decreasing nature.

For a function like \( f(x) = x + \sin x - b \), the derivative, represented by \( f'(x) \), is calculated using rules of differentiation. Let's break it down:

• \( \frac{dx}{dx} \) is the derivative of \( x \) with respect to \( x \), which is always 1.
• \( \frac{d(\sin x)}{dx} \) is the derivative of \( \sin x \), which is \( \cos x \) due to the trigonometric functions differentiation rules.
• \( \frac{db}{dx} \) is the derivative of a constant \( b \), which is 0 because constants don't change with respect to \( x \).

Therefore, combining these results we obtain the derivative as \( f'(x) = 1 + \cos x \). Since \( \cos x \) can range from -1 to 1, we know that the smallest possible value for \( f'(x) \) is 0, which tells us that \( f(x) \) is at least non-decreasing everywhere, and can only increase, but never decrease.

Increasing Function Analysis

An increasing function is one where the output value rises as the input value increases. This analysis can help us determine whether a function's graph is climbing or falling as we move along it from left to right.

To analyze whether \( f(x) = x + \sin x - b \) is an increasing function, we look at the sign of its derivative \( f'(x) = 1 + \cos x \). A positive derivative means the function's rate of change is positive and thus the function is increasing. In our case, considering \( 1 + \cos x \geq 0 \), we can conclude that \( f(x) \) must be increasing for all \( x \).

This is crucial in understanding the number of roots a function may have. If a function is ever-increasing, it can cross the x-axis only once, since after crossing, it continues to ascend, making further intersections impossible. In the context of our exercise, this implies that there can be at most one positive root for \( f(x) \) when \( b > 0 \), and none when \( b < 0 \), as an increasing function can't dip back below the x-axis after initially starting above it (for \( b < 0 \), \( f(0) > 0 \)).

Trigonometric Functions Differentiation

Differentiation of trigonometric functions is part of the toolkit for analyzing the behavior of equations involving sine, cosine, and other trigonometric terms.

Considering our function \( f(x) = x + \sin x - b \), the derivative of the \( \sin x \) portion is \( \cos x \), which stems from a fundamental rule of trigonometric differentiation. Fields such as physics, engineering, and even economics frequently use trigonometric functions, and thus understanding their rates of change is invaluable.

For instance, in our exercise, the range of the cosine function dictates that \( f'(x) \) will never be negative since it's being added to 1. Hence, \( f'(x) \) is always non-negative, reaffirming that \( f(x) \) is an increasing function. This property guides us to make accurate predictions about the function's graph and its intersections—or lack thereof—with the x-axis, leading to conclusions about the existence of roots.