Problem 64 - University Calculus: Early Transcendentals

Question

You have been asked to determine whether the function \(f(x)=\) \(3+4 \cos x+\cos 2 x\) is ever negative. a. Explain why you need to consider values of \(x\) only in the interval \([0,2 \pi]\) b. Is \(f\) ever negative? Explain.

Step-by-Step Solution

Answer

f(x) is never negative; it is 0 at its minimum.

Step 1: Understanding the Given Function

The function given is \(f(x) = 3 + 4\cos x + \cos 2x\). The task is to find if this function ever takes negative values.

Step 2: Using Trigonometric Identity

We know that \(\cos 2x = 2\cos^2 x - 1\). Substitute this identity into the function to simplify it: \[f(x) = 3 + 4\cos x + (2\cos^2 x - 1)\]which simplifies to \[f(x) = 2\cos^2 x + 4\cos x + 2\].

Step 3: Determine the Minimum Value of the Quadratic Expression

The expression is a quadratic in \(\cos x\): \[f(x) = 2(\cos^2 x) + 4(\cos x) + 2\].To find the minimum value, calculate the vertex of the quadratic, given by \(x = -\frac{b}{2a}\). Here, \(a = 2\), \(b = 4\). Therefore, \(\cos x = -\frac{4}{4} = -1\).

Step 4: Evaluating the Function at the Critical Point

Evaluate \(f(x)\) when \(\cos x = -1\):\[f(x) = 2(-1)^2 + 4(-1) + 2 = 2 - 4 + 2 = 0\].Thus, the minimum value of \(f(x)\) is 0.

Step 5: Explain the Consideration of the Interval [0, 2π]

The primary reason to consider the interval \([0, 2\pi]\) is because trigonometric functions like \(\cos x\) and \(\cos 2x\) are periodic with a period of \(2\pi\). Therefore, all possible values of \(f(x)\) will be included within a single cycle of \(\cos x\) from \(0\) to \(2\pi\).

Step 6: Conclusion about the Function Value

Since the minimum value of \(f(x)\) is 0, and \(f(x)\) does not take values below 0, the function \(f(x) = 3 + 4\cos x + \cos 2x\) is never negative.

Key Concepts

Trigonometric identities

Understanding trigonometric identities is crucial when analyzing trigonometric functions, as they allow us to simplify complex expressions. In the given problem, recognizing the identity for the double angle of cosine greatly simplifies the function's analysis. The double angle identity is expressed as

• \(\cos 2x = 2\cos^2 x - 1\)

This identity enables us to rewrite expressions involving \(\cos 2x\) into terms of \(\cos x\), simplifying the task of analyzing and manipulating the function.
By substituting this identity into the original function, we've changed \(f(x) = 3 + 4\cos x + \cos 2x\) into a simpler quadratic expression in terms of \(\cos x\):

• \(f(x) = 2\cos^2 x + 4\cos x + 2\)

This change is possible due to the powerful substitutions allowed by trigonometric identities, making it easier to analyze the function's behavior.

Function analysis

Function analysis often involves determining critical points and evaluating function behavior across intervals. In this problem, once the expression was rewritten, we were able to treat it as a quadratic function with respect to \(\cos x\). This is understood as\[f(x) = 2(\cos^2 x) + 4(\cos x) + 2\]To find the critical point, we calculate the vertex of the quadratic using the standard formula for the vertex, \(x = -\frac{b}{2a}\). Here, \(a = 2\), and \(b = 4\), leading to:

• \(\cos x = -\frac{4}{4} = -1\)

Evaluating the function at this critical point results in:\[f(x) = 2(-1)^2 + 4(-1) + 2 = 0\] This indicates that the minimum value of the function within a cycle is zero, showing how function analysis can reveal crucial insights about when the function reaches its extremum values.

Periodicity of trigonometric functions

Trigonometric functions, like cosine, are naturally periodic, repeating their values in regular intervals. This periodic nature is why we focus primarily on the interval \[0, 2\pi\]\ for our analysis. Within one full cycle of its period, from 0 to \(2\pi\), the function exhibits all possible values it can take.
The function \(f(x) = 3 + 4\cos x + \cos 2x\) is constructed from periodic components \(\cos x\) and \(\cos 2x\). Both \(\cos x\) and \(\cos 2x\) share a period of \(2\pi\) , meaning all significant behavior of \(f(x)\) is captured within this interval.

• The property of periodicity ensures that we do not miss any potential minimum or maximum values of the function because after each interval of \(2\pi\), the function values repeat. Since the minimum found within one period is zero, \(f(x)\) does not become negative anywhere in its domain.

Using periodicity, we limited our examination to one interval, confident it represents the entire behavior of our trigonometric function.