MHB Finding the domain of this function.

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The function f(x) = 1/√(x² - 2x cos(θ) + 4) has a domain that varies based on the value of θ within [0, π]. For θ = π/2, x can take any real value, while for θ = 0 and θ = π, x cannot equal 2 and -2, respectively. For values of θ between 0 and π, the expression under the square root is always positive, indicating that the domain is all real numbers. The discussion highlights the importance of understanding the behavior of the quadratic expression in relation to θ. Ultimately, the domain of the function is confirmed to be all real numbers, R.
cbarker1
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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
 
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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.
 
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