Finding the domain of this function.

[cbarker1]

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

 

[skeeter]

$f(x)=\frac{1}{\sqrt{x^2-2x\cos(\theta)+4}}$ 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

 

[Prove It]

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 $\mathbf{R}$.

 

[skeeter]

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.

 

[Prove It]

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.

 

[skeeter]

No upset ... it happens.