Finding a domain for a function

In summary: So the domain of f(x) is either the whole real line (if \cos^2 \theta \neq 1) or a single point (if \cos^2 \theta =1).In summary, the domain of the given function ##f(x)=\frac{1}{\sqrt{x^2-4x\cos(\theta)+4}}## and ##\theta\in[0,\pi]## is either the whole real line (if ##\cos^2 \theta \neq 1##) or a single point (if ##\cos^2 \theta =1##).
  • #1
cbarker1
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Homework Statement
Find the domain of this Function:
Relevant Equations
Quadratic Formula, Trig. Identity.
I am having some trouble find the domain with this function: ##f(x)=\frac{1}{\sqrt{x^2-4x\cos(\theta)+4}}## and ##\theta\in[0,\pi]##.I know that the denominator needs to be greater than 0. So ##\sqrt{x^2-4x\cos(\theta)+4}>0##. I squared both side of the inequality, ##x^2-4x\cos(\theta)+4>0##. Then I use the quadratic formula in terms of x: ##x>\frac{4\cos(\theta)\pm\sqrt{16(\cos(\theta)^2-16}}{2}##. With some simplification and using the trig. identity, I got ##x> 2\cos(\theta)\pm 2\sin(\theta)##. But I do not know how to proceed from here.

Thanks,
cbarker1
 
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  • #2
cbarker1 said:
Then I use the quadratic formula in terms of x
Why ? That gives you the values of x where ##x^2-4x\cos(\theta)+4=0## -- if any !
And it is immediately clear that there are no such points if ##\cos\theta\ne 1##.
cbarker1 said:
With some simplification and using the trig. identity
I wonder how you do that ?

##\ ##
 
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  • #3
Because the original function have a square root in the denominator. I can pull out the ##\sqrt{16}\sqrt{\cos(\theta)^2-1}=4\sqrt{\cos(\theta)^2-1}##. I realized the trig identity for ##\sqrt{\cos(\theta)-1}## does not exist. So then, the only values will not work is when ##\sqrt{\cos(\theta)-1}>0##.
 
  • #4
cbarker1 said:
I am having some trouble find the domain with this function:
##f(x)=\frac{1}{\sqrt{x^2-4x\cos(\theta)+4}}## and ##\theta\in[0,\pi]##.
It seems to me that f is a function of two variables, x and ##\theta##, not just x. So then the domain would be a two-dimensional region in the ##x-\theta## plane.
 
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  • #5
If the thing in the square root is zero, the function is clearly not defined. If the thing in the square root is negative, it's also not defined. How is the region where it's negative related to the region where it's zero?
 
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  • #6
cbarker1 said:
Because the original function have a square root in the denominator. I can pull out the ##\sqrt{16}\sqrt{\cos(\theta)^2-1}=4\sqrt{\cos(\theta)^2-1}##. I realized the trig identity for ##\sqrt{\cos(\theta)-1}## does not exist. So then, the only values will not work is when ##\sqrt{\cos(\theta)-1}>0##.
Doesn't make sense. ##\cos\theta - 1 ## is never ##> 0## !

If 'it doesn't work' that means the original function can be evaluated.

How about writing ## \ x^2-4x\cos\theta+4 \ ## as ##(x-2\cos\theta)^2+4 (1-\cos^2\theta)## ?
The first one is ##\ge \ (x-2)^2 \ ## and the second one ##\ge 0 ## and you only have one point to investigate.

Mark44 said:
It seems to me that f is a function of two variables, x and ##\theta##, not just x. So then the domain would be a two-dimensional region in the ##x-\theta## plane.
##\theta\in[0,\pi]## is given.
 
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  • #7
BvU said:
##\theta\in[0,\pi]## is given.
Right, but should ##\theta## be considered a variable, in which case we have f being a function of two variables? Or should ##\theta## be considered a parameter, an unspecified, but fixed, value? In either case, the value of f depends on the choice of both x and ##\theta##.
 
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  • #8
cbarker1 said:
Homework Statement:: Find the domain of this Function:
Relevant Equations:: Quadratic Formula, Trig. Identity.

I am having some trouble find the domain with this function: ##f(x)=\frac{1}{\sqrt{x^2-4x\cos(\theta)+4}}## and ##\theta\in[0,\pi]##.I know that the denominator needs to be greater than 0. So ##\sqrt{x^2-4x\cos(\theta)+4}>0##. I squared both side of the inequality, ##x^2-4x\cos(\theta)+4>0##. Then I use the quadratic formula in terms of x: ##x>\frac{4\cos(\theta)\pm\sqrt{16(\cos(\theta)^2-16}}{2}##. With some simplification and using the trig. identity, I got ##x> 2\cos(\theta)\pm 2\sin(\theta)##.

Look again at the discriminant: [tex]
16\cos^2\theta -16 = 16(\cos^2 \theta - 1) \leq 0[/tex] so there are no real roots unless [itex]\cos^2 \theta = 1[/itex], when there is exactly one real root.
 
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Related to Finding a domain for a function

What is a domain?

A domain is the set of all possible input values for a function. It is the independent variable in a function and is typically represented by the letter "x".

Why is finding a domain important?

Finding a domain is important because it helps to determine the range of values that the function can output. It also ensures that the function is well-defined and avoids any potential errors or undefined values.

How do I find the domain of a function?

To find the domain of a function, you need to identify any restrictions on the input values. This can be done by looking for excluded values such as division by zero or negative numbers under a square root.

Can a function have more than one domain?

No, a function can only have one domain. The domain is a unique set of input values that correspond to unique output values. However, a function can have multiple inputs that result in the same output.

What happens if I choose the wrong domain for a function?

If you choose the wrong domain for a function, it can result in incorrect or undefined output values. This can lead to errors in calculations and make the function invalid. It is important to carefully consider the domain when working with functions.

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