Problem 75
Question
A function \(f\) is given. (a) Is \(f\) even, odd, or neither? (b) Find the \(x\) -intercepts of the graph of \(f\) (c) Graph \(f\) in an appropriate viewing rectangle. (d) Describe the behavior of the function as \(x \rightarrow \pm \infty\). (e) Notice that \(f(x)\) is not defined when \(x=0 .\) What happens as \(x\) approaches \(0 ?\) $$f(x)=\frac{1-\cos x}{x}$$
Step-by-Step Solution
Verified Answer
(a) Odd; (b) \( x = \pm 2\pi, \pm 4\pi, ... \); (c) Graph reflecting odd symmetry, undefined at 0; (d) Approaches 0 as \( x \rightarrow \pm \infty \); (e) Approaches 0 as \( x \rightarrow 0 \).
1Step 1: Test for Symmetry
To determine whether the function is even or odd, compute \( f(-x) \). Given \( f(x) = \frac{1 - \cos x}{x} \), we have: \[ f(-x) = \frac{1 - \cos(-x)}{-x} = \frac{1 - \cos x}{-x} = -f(x) \]Since \( f(-x) = -f(x) \), the function is odd.
2Step 2: Find x-intercepts
The x-intercepts of a function occur where \( f(x) = 0 \). For \( f(x) = \frac{1 - \cos x}{x} \), set the numerator equal to zero: \[ 1 - \cos x = 0 \]\[ \cos x = 1 \]This occurs at \( x = 2\pi k \), for integer \(k\). However, considering the undefined behavior at \( x = 0 \), the x-intercepts are \( x = \pm 2\pi, \pm 4\pi, \pm 6\pi, ... \) etc., excluding 0.
3Step 3: Graph the Function
Graph \( f(x) \) using software or by hand, keeping in mind it is undefined at \( x = 0 \). The graph reflects odd symmetry with discontinuity at the origin. Choose a viewing window that includes multiple periods of the cosine function, such as \([-4\pi, 4\pi]\) for x and \([-0.5, 0.5]\) for y.
4Step 4: Behavior as \(x \rightarrow \pm \infty\)
As \( x \rightarrow \pm \infty \), the behavior of \( f(x) \) is influenced by \( \frac{1}{x} \) tending towards zero, and since \(|\cos x|\leq 1\), overall the function tends towards zero. Therefore, \( \lim_{x \to \pm \infty} f(x) = 0 \).
5Step 5: Analyze Near \(x = 0\)
As \( x \rightarrow 0 \), the denominator approaches zero causing \( f(x) \) to become undefined. Using L'Hôpital's rule: \[ \lim_{x \to 0} \frac{1-\cos x}{x} = \lim_{x \to 0} \frac{\sin x}{1} = 0 \]Thus, \( f(x) \) approaches zero, even though it is technically undefined at \( x = 0 \).
Key Concepts
Even and Odd FunctionsX-InterceptsLimits and Behavior at InfinityL'Hôpital's Rule
Even and Odd Functions
When we discuss whether a function is even, odd, or neither, we look at the symmetry of the function. For a function to be considered even, it must satisfy the condition that \( f(-x) = f(x) \). This suggests that the graph of the function is symmetric with respect to the y-axis. On the other hand, for a function to be odd, it must meet the condition that \( f(-x) = -f(x) \). This translates to symmetry about the origin.
If a function doesn't satisfy either of these conditions, it is neither even nor odd. In the given function \( f(x) = \frac{1 - \cos x}{x} \), when we substitute \(-x\) in place of \(x\), we find that \( f(-x) = -f(x) \). Hence, the function is odd, displaying that lovely characteristic of origin symmetry.
Understanding these symmetries helps simplify many problems, especially when analyzing the graph of the function.
If a function doesn't satisfy either of these conditions, it is neither even nor odd. In the given function \( f(x) = \frac{1 - \cos x}{x} \), when we substitute \(-x\) in place of \(x\), we find that \( f(-x) = -f(x) \). Hence, the function is odd, displaying that lovely characteristic of origin symmetry.
Understanding these symmetries helps simplify many problems, especially when analyzing the graph of the function.
X-Intercepts
X-intercepts are the points where the function crosses the x-axis, meaning \( f(x) = 0 \). To find the x-intercepts of a function, set the function equal to zero and solve for \(x\). For the function \( f(x) = \frac{1 - \cos x}{x} \), the numerator must be zero for the function to have x-intercepts. This occurs when \( 1 - \cos x = 0 \), leading us to \( \cos x = 1 \).
The cosine function equals 1 at integer multiples of \(2\pi\) (like \(0, \pm 2\pi, \pm 4\pi, \ldots\)). However, since \(f(x)\) is undefined at \(x = 0\), we need to exclude it from our intercepts. Therefore, the x-intercepts are \(x = \pm 2\pi, \pm 4\pi, \pm 6\pi, \ldots\), excluding zero.
These intercepts are scattered along the x-axis at regular intervals, aligning with the periodic nature of the cosine function.
The cosine function equals 1 at integer multiples of \(2\pi\) (like \(0, \pm 2\pi, \pm 4\pi, \ldots\)). However, since \(f(x)\) is undefined at \(x = 0\), we need to exclude it from our intercepts. Therefore, the x-intercepts are \(x = \pm 2\pi, \pm 4\pi, \pm 6\pi, \ldots\), excluding zero.
These intercepts are scattered along the x-axis at regular intervals, aligning with the periodic nature of the cosine function.
Limits and Behavior at Infinity
To understand the behavior of a function as \(x\) approaches infinity (\(+\infty\)) or negative infinity (\(-\infty\)), we analyze the limits. The function \( f(x) = \frac{1 - \cos x}{x} \) offers an interesting case. As \(x\) becomes very large or very small (negatively large), the term \( \frac{1}{x} \) in the function acts like a reducing factor because it approaches zero.
Since \( \cos x \) oscillates between -1 and 1, its effect diminishes as \(x\) grows in magnitude. Therefore, the function \(f(x)\) tends towards zero as \(x\) approaches infinity or negative infinity, giving us \( \lim_{x \to \pm \infty} f(x) = 0 \).
Since \( \cos x \) oscillates between -1 and 1, its effect diminishes as \(x\) grows in magnitude. Therefore, the function \(f(x)\) tends towards zero as \(x\) approaches infinity or negative infinity, giving us \( \lim_{x \to \pm \infty} f(x) = 0 \).
- This tells us that the function levels out and becomes negligible as \(x\) strides far off in either direction.
L'Hôpital's Rule
L'Hôpital's Rule is a mighty tool in calculus, used to evaluate limits of indeterminate forms. Typically, forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) need this approach. For our function \( f(x) = \frac{1 - \cos x}{x} \), as \(x\) approaches zero, both the numerator and denominator tend to zero, setting up an indeterminate \(\frac{0}{0}\) case.
L'Hôpital's Rule states that if the limit of \( \frac{u(x)}{v(x)} \) as \(x\) approaches some value is indeterminate, then \( \lim_{x \to a} \frac{u(x)}{v(x)} \) is \( \lim_{x \to a} \frac{u'(x)}{v'(x)} \), given this limit exists. Differentiating top and bottom of \(f(x)\), we get \( \lim_{x \to 0} \frac{\sin x}{1} \).
Now, since \( \lim_{x \to 0} \sin x = 0 \), we find that \( \lim_{x \to 0} f(x) = 0 \).
L'Hôpital's Rule states that if the limit of \( \frac{u(x)}{v(x)} \) as \(x\) approaches some value is indeterminate, then \( \lim_{x \to a} \frac{u(x)}{v(x)} \) is \( \lim_{x \to a} \frac{u'(x)}{v'(x)} \), given this limit exists. Differentiating top and bottom of \(f(x)\), we get \( \lim_{x \to 0} \frac{\sin x}{1} \).
Now, since \( \lim_{x \to 0} \sin x = 0 \), we find that \( \lim_{x \to 0} f(x) = 0 \).
- This clever rule hands us the behavior of functions at otherwise tricky points.
Other exercises in this chapter
Problem 74
Find all solutions of the equation that lie in the interval \([0, \pi] .\) State each answer correct to two decimal places. $$\cos x=x$$
View solution Problem 75
Determine whether the function is even, odd, or neither. $$f(x)=\sin x \cos x$$
View solution Problem 76
Determine whether the function is even, odd, or neither. $$f(x)=\sin x+\cos x$$
View solution Problem 76
A function \(f\) is given. (a) Is \(f\) even, odd, or neither? (b) Find the \(x\) -intercepts of the graph of \(f\) (c) Graph \(f\) in an appropriate viewing re
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