# Finding the domain of this function.

• MHB
• cbarker1
In summary, the function $f(x)=\frac{1}{\sqrt{x^2-2x\cos(\theta)+4}}$ is defined for all real values of $x$ and $\theta$ for the given domain of $\theta\in[0,\pi]$. This is due to the fact that the denominator, $x^2-2x\cos(\theta)+4$, is always greater than or equal to 3, ensuring that the square root is always defined.
cbarker1
Gold Member
MHB
Dear Everyone,

I am having some trouble find the domain with this function: $f(x)=\frac{1}{\sqrt{x^2-2x\cos(\theta)+4}}$ and $\theta\in[0,\pi]$.

My attempt:

I know that the denominator needs to be greater than 0. So $\sqrt{x^2-2x\cos(\theta)+2}>0$. I squared both side of the inequality. Then I use the quadratic formula in terms of x: $x>\frac{2\cos(\theta)\pm\sqrt{4(\cos(\theta)^2-4}}{2}$. With some simplification and using the trig. identities, I got $x> \cos(\theta)\pm \sin(\theta)$. But I do not know how to proceed from here. Thanks,
Cbarker1

Cbarker1 said:
$f(x)=\frac{1}{\sqrt{x^2-2x\cos(\theta)+4}}$ and $\theta\in[0,\pi]$.

the domain depends of the value of $\theta$ ...

$\theta = \dfrac{\pi}{2} \implies f(x) = \dfrac{1}{\sqrt{x^2+4}}$

... $x$ can be any real value

$\theta = 0 \implies f(x) = \dfrac{1}{\sqrt{(x-2)^2}} = \dfrac{1}{|x-2|}$

... $x \ne 2$

$\theta = \pi \implies f(x) = \dfrac{1}{\sqrt{(x+2)^2}} = \dfrac{1}{|x+2|}$

... $x \ne -2$

for $0 < \theta < \dfrac{\pi}{2}$

$0 < \cos{\theta} < 1 \implies 0 < 2x\cos{\theta} < 2x$ for $x > 0 \implies x^2-2x\cos{\theta}+4 > x^2 - 2x + 4 \ge 0$

$0 < \cos{\theta} < 1 \implies 2x < 2x\cos{\theta} < 0$ for $x < 0 \implies x^2-2x\cos{\theta}+4 > 0$

... $x$ can be any real value

for $\dfrac{\pi}{2}< \theta < \pi$

$-1 < \cos{\theta} < 0 \implies -2x < 2x\cos{\theta} < 0$ for $x > 0 \implies x^2-2x\cos{\theta}+4 > 0$

$-1 < \cos{\theta} < 0 \implies 0 < 2x\cos{\theta} < -2x$ for $x < 0 \implies x^2-2x\cos{\theta}+4 > x^2-2x+4 \ge 0$

... $x$ can be any real value

Cbarker1 said:
Dear Everyone,

I am having some trouble find the domain with this function: $f(x)=\frac{1}{\sqrt{x^2-2x\cos(\theta)+4}}$ and $\theta\in[0,\pi]$.

My attempt:

I know that the denominator needs to be greater than 0. So $\sqrt{x^2-2x\cos(\theta)+2}>0$. I squared both side of the inequality. Then I use the quadratic formula in terms of x: $x>\frac{2\cos(\theta)\pm\sqrt{4(\cos(\theta)^2-4}}{2}$. With some simplification and using the trig. identities, I got $x> \cos(\theta)\pm \sin(\theta)$. But I do not know how to proceed from here.Thanks,
Cbarker1

\displaystyle \begin{align*} x^2 - 2\cos{ \left( \theta \right) } \, x + 4 &> 0 \\ x^2 - 2\cos{ \left( \theta \right) }\, x + \left[ -\cos{ \left( \theta \right) } \right] ^2 - \left[ -\cos{ \left( \theta \right) } \right] ^2 + 4 &> 0 \\ \left[ x - \cos{ \left( \theta \right) } \right] ^2 - \cos^2{ \left( \theta \right) } + 4 &> 0 \end{align*}

Note that $\displaystyle \left[ x - \cos{ \left( \theta \right) } \right] ^2 \geq 0$ for all $x, \theta$ and also

\displaystyle \begin{align*} 0 \leq \cos^2{ \left( \theta \right) } &\leq 1 \textrm{ for all } \theta, \textrm{ so } \\ -1 \leq -\cos^2{ \left( \theta \right) } &\leq 0 \\ 3 \leq -\cos^2{ \left( \theta \right) } + 4 &\leq 4 \end{align*}

so that means $\displaystyle \left[ x - \cos{ \left( \theta \right) } \right] ^2 - \cos^2{ \left( \theta \right) } + 4 \geq 3$ for all $x, \theta$.

Since the square root amount is never any less than 3, it's going to always be defined. The domain is $\mathbf{R}$.

Prove It said:
Since the square root amount is never any less than 3, it's going to always be defined. The domain is $\mathbf{R}$.

that is correct ... I was looking at $x^2 \pm 2x +4$ as if it were $x^2 \pm 4x +4$ when I factored.

mea culpa.

skeeter said:
that is correct ... I was looking at $x^2 \pm 2x +4$ as if it were $x^2 \pm 4x +4$ when I factored.

mea culpa.

Sorry Skeeter, didn't mean to upset you. I just liked my strategy so wanted to share it hahaha.

No upset ... it happens.

## What is the domain of a function?

The domain of a function is the set of all possible input values for which the function is defined. In other words, it is the set of all values that can be plugged into the function to produce a valid output.

## Why is it important to find the domain of a function?

Finding the domain of a function is important because it helps us understand the behavior of the function and identify any restrictions or limitations on its input values. It also ensures that we are working with valid inputs and can accurately interpret the results of the function.

## How do you find the domain of a function algebraically?

To find the domain of a function algebraically, you need to identify any restrictions on the input values. This can include restrictions on the denominator of a fraction, the radicand of a square root, or the variable in the argument of a logarithmic or trigonometric function. Once these restrictions are identified, the domain is the set of all real numbers that satisfy these restrictions.

## Can the domain of a function ever be all real numbers?

No, the domain of a function can never be all real numbers. There are certain mathematical operations, such as division by zero, that are undefined and therefore cannot be included in the domain. Additionally, some functions have natural limitations on their input values, such as a square root function only accepting non-negative inputs.

## What are some common mistakes when finding the domain of a function?

Some common mistakes when finding the domain of a function include forgetting to check for restrictions on the input values, incorrectly identifying the domain as all real numbers, and not considering the context of the problem. It is important to carefully analyze the function and its limitations to accurately determine its domain.

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