Evaluating the Function at the Critical Points

Oct 13, 2025 by magicalceylontours.com 0 views

Analyzing the Given Function: $f(x)=\frac{1}{1+\cos x}$

In this article, we will be analyzing the given function $f(x)=\frac{1}{1+\cos x}$ and determining the value of $x$ for which $f$ is undefined, as well as the minimum value of $f$ . The function $f(x)$ is a trigonometric function that involves the cosine function. We will use various mathematical techniques to analyze and understand the behavior of this function.

The Value of $x$ for Which $f$ is Undefined

To determine the value of $x$ for which $f$ is undefined, we need to find the values of $x$ that make the denominator of the function equal to zero. The denominator of the function is $1+\cos x$ . We know that the cosine function has a range of $[-1, 1]$ . Therefore, the denominator $1+\cos x$ will be equal to zero when $\cos x = -1$ .

The cosine function is equal to $-1$ at odd multiples of $180^\circ$ . Therefore, the value of $x$ for which $f$ is undefined is given by:

$x = (2n + 1) \cdot 180^\circ$

where $n$ is an integer.

The Minimum Value of $f$

To determine the minimum value of $f$ , we need to find the critical points of the function. The critical points of the function are the values of $x$ for which the derivative of the function is equal to zero or undefined.

The derivative of the function $f(x)$ is given by:

$f'(x) = \frac{-\sin x}{(1+\cos x)^2}$

We can see that the derivative is undefined when the denominator is equal to zero, which occurs when $\cos x = -1$ . Therefore, the critical points of the function are the same as the values of $x$ for which $f$ is undefined.

To find the minimum value of $f$ , we need to evaluate the function at the critical points. We can do this by substituting the values of $x$ into the function.

Evaluating the Function at the Critical Points

We can evaluate the function at the critical points by substituting the values of $x$ into the function. We get:

$f(0^\circ) = \frac{1}{1+\cos 0^\circ} = \frac{1}{2}$

$f(180^\circ) = \frac{1}{1+\cos 180^\circ} = \frac{1}{0}$

$f(360^\circ) = \frac{1}{1+\cos 360^\circ} = \frac{1}{2}$

We can see that the function is undefined at $x = 180^\circ$ . Therefore, the minimum value of $f$ occurs at $x = 0^\circ$ or $x = 360^\circ$ .

In conclusion, we have analyzed the given function $f(x)=\frac{1}{1+\cos x}$ and determined the value of $x$ for which $f$ is undefined, as well as the minimum value of $f$ . We found that the value of $x$ for which $f$ is undefined is given by $x = (2n + 1) \cdot 180^\circ$ , and the minimum value of $f$ occurs at $x = 0^\circ$ or $x = 360^\circ$ .

[1] Calculus, James Stewart, 7th edition
[2] Trigonometry, Michael Corral, 2nd edition

Proof of the Derivative

To prove the derivative of the function $f(x)$ , we can use the quotient rule of differentiation.

Let $u(x) = 1$ and $v(x) = 1 + \cos x$ . Then, we have:

$f(x) = \frac{u(x)}{v(x)}$

The derivative of $f(x)$ is given by:

$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{v(x)^2}$

We can see that $u'(x) = 0$ and $v'(x) = -\sin x$ . Therefore, we have:

$f'(x) = \frac{0 \cdot (1 + \cos x) - 1 \cdot (-\sin x)}{(1 + \cos x)^2}$

Simplifying the expression, we get:

$f'(x) = \frac{\sin x}{(1 + \cos x)^2}$

This is the derivative of the function $f(x)$ .

The derivative of the function $f(x)$ is undefined when the denominator is equal to zero, which occurs when $\cos x = -1$ . Therefore, the critical points of the function are the same as the values of $x$ for which $f$ is undefined.
Q&A: Analyzing the Given Function $f(x)=\frac{1}{1+\cos x}$

In our previous article, we analyzed the given function $f(x)=\frac{1}{1+\cos x}$ and determined the value of $x$ for which $f$ is undefined, as well as the minimum value of $f$ . In this article, we will answer some frequently asked questions related to the function.

Q: What is the domain of the function $f(x)$ ?

A: The domain of the function $f(x)$ is all real numbers except for the values of $x$ that make the denominator equal to zero. These values are given by $x = (2n + 1) \cdot 180^\circ$ , where $n$ is an integer.

Q: What is the range of the function $f(x)$ ?

A: The range of the function $f(x)$ is all real numbers except for zero. This is because the function is undefined when the denominator is equal to zero, and the denominator is never equal to zero when the function is equal to zero.

Q: How do you find the critical points of the function $f(x)$ ?

A: To find the critical points of the function $f(x)$ , you need to find the values of $x$ for which the derivative of the function is equal to zero or undefined. The derivative of the function is given by $f'(x) = \frac{-\sin x}{(1+\cos x)^2}$ . The critical points are the values of $x$ that make the derivative equal to zero or undefined.

Q: What is the minimum value of the function $f(x)$ ?

A: The minimum value of the function $f(x)$ occurs at the critical points. To find the minimum value, you need to evaluate the function at the critical points. The minimum value of the function is $\frac{1}{2}$ , which occurs at $x = 0^\circ$ or $x = 360^\circ$ .

Q: How do you graph the function $f(x)$ ?

A: To graph the function $f(x)$ , you need to plot the values of the function at various points. You can use a graphing calculator or a computer program to graph the function. The graph of the function will show the values of the function at various points, including the critical points.

Q: What is the significance of the function $f(x)$ ?

A: The function $f(x)$ is a trigonometric function that involves the cosine function. It is used in various mathematical and scientific applications, including physics, engineering, and computer science. The function is also used in the study of trigonometry and calculus.

In conclusion, we have answered some frequently asked questions related to the function $f(x)=\frac{1}{1+\cos x}$ . We have discussed the domain and range of the function, the critical points, the minimum value, and the graph of the function. We hope that this article has been helpful in understanding the function and its applications.

[1] Calculus, James Stewart, 7th edition
[2] Trigonometry, Michael Corral, 2nd edition