Finding a domain for a function

[cbarker1]

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

 

[BvU]

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 $\cos\theta\ne 1$.

With some simplification and using the trig. identity

I wonder how you do that ?

$\$

 

[cbarker1]

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 $\sqrt{\cos(\theta)-1}>0$.

 

[Mark44]

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.

 

[Office_Shredder]

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?

 

[BvU]

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.

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.

 

[Mark44]

$\theta\in[0,\pi]$ 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 $\theta$.

 

[pasmith]

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: $$16\cos^2\theta -16 = 16(\cos^2 \theta - 1) \leq 0$$ so there are no real roots unless $\cos^2 \theta = 1$, when there is exactly one real root.