Domain of functions involving arcsine or arccosine











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I have two rather simple functions. I don't know how to find that the dominator is different than 0. Find the domain of functions:



Function 1:



$f(x) = frac{4-x}{arcsinfrac{x}{4}}$



Assumption 1:



$-1leq arcsinfrac{x}{4} leq 1$



$-1leq frac{x}{4} leq 1$



$-4leq x leq 4$



$xin<4;4>$



Assumption 2:



$arcsinfrac{x}{4} neq 0$



Here, I have no idea how to proceed further with assumption 2.



Function 2:



$f(x) = frac{sqrt{2x-1}}{2+arccosfrac{x+1}{4}}$



Assumption 1:



$sqrt{2x-1} geq 0$



$2x geq 1$



$x geq frac{1}{2}$



Assumption 2:



$ -1 leq arccos frac{x+1}{4} leq 1 $



$ -1 leq frac{x+1}{4} leq 1 $



$ -4 leq x + 1 leq 4 $



$ -5 leq x leq 3 $



Assumption 3:



$ 2 + arccosfrac{x+1}{4} neq 0 $



Here once again, no idea how to proceed further.



Would anyone be able to help me understand these examples, how to find that the denominator is not equal to $0$, when there's arccos or arcsin in the denominator? Thanks! I do hope my tags are okay...










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  • Use en.wikipedia.org/wiki/…
    – lab bhattacharjee
    2 days ago















up vote
2
down vote

favorite












I have two rather simple functions. I don't know how to find that the dominator is different than 0. Find the domain of functions:



Function 1:



$f(x) = frac{4-x}{arcsinfrac{x}{4}}$



Assumption 1:



$-1leq arcsinfrac{x}{4} leq 1$



$-1leq frac{x}{4} leq 1$



$-4leq x leq 4$



$xin<4;4>$



Assumption 2:



$arcsinfrac{x}{4} neq 0$



Here, I have no idea how to proceed further with assumption 2.



Function 2:



$f(x) = frac{sqrt{2x-1}}{2+arccosfrac{x+1}{4}}$



Assumption 1:



$sqrt{2x-1} geq 0$



$2x geq 1$



$x geq frac{1}{2}$



Assumption 2:



$ -1 leq arccos frac{x+1}{4} leq 1 $



$ -1 leq frac{x+1}{4} leq 1 $



$ -4 leq x + 1 leq 4 $



$ -5 leq x leq 3 $



Assumption 3:



$ 2 + arccosfrac{x+1}{4} neq 0 $



Here once again, no idea how to proceed further.



Would anyone be able to help me understand these examples, how to find that the denominator is not equal to $0$, when there's arccos or arcsin in the denominator? Thanks! I do hope my tags are okay...










share|cite|improve this question









New contributor




weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Use en.wikipedia.org/wiki/…
    – lab bhattacharjee
    2 days ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











I have two rather simple functions. I don't know how to find that the dominator is different than 0. Find the domain of functions:



Function 1:



$f(x) = frac{4-x}{arcsinfrac{x}{4}}$



Assumption 1:



$-1leq arcsinfrac{x}{4} leq 1$



$-1leq frac{x}{4} leq 1$



$-4leq x leq 4$



$xin<4;4>$



Assumption 2:



$arcsinfrac{x}{4} neq 0$



Here, I have no idea how to proceed further with assumption 2.



Function 2:



$f(x) = frac{sqrt{2x-1}}{2+arccosfrac{x+1}{4}}$



Assumption 1:



$sqrt{2x-1} geq 0$



$2x geq 1$



$x geq frac{1}{2}$



Assumption 2:



$ -1 leq arccos frac{x+1}{4} leq 1 $



$ -1 leq frac{x+1}{4} leq 1 $



$ -4 leq x + 1 leq 4 $



$ -5 leq x leq 3 $



Assumption 3:



$ 2 + arccosfrac{x+1}{4} neq 0 $



Here once again, no idea how to proceed further.



Would anyone be able to help me understand these examples, how to find that the denominator is not equal to $0$, when there's arccos or arcsin in the denominator? Thanks! I do hope my tags are okay...










share|cite|improve this question









New contributor




weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I have two rather simple functions. I don't know how to find that the dominator is different than 0. Find the domain of functions:



Function 1:



$f(x) = frac{4-x}{arcsinfrac{x}{4}}$



Assumption 1:



$-1leq arcsinfrac{x}{4} leq 1$



$-1leq frac{x}{4} leq 1$



$-4leq x leq 4$



$xin<4;4>$



Assumption 2:



$arcsinfrac{x}{4} neq 0$



Here, I have no idea how to proceed further with assumption 2.



Function 2:



$f(x) = frac{sqrt{2x-1}}{2+arccosfrac{x+1}{4}}$



Assumption 1:



$sqrt{2x-1} geq 0$



$2x geq 1$



$x geq frac{1}{2}$



Assumption 2:



$ -1 leq arccos frac{x+1}{4} leq 1 $



$ -1 leq frac{x+1}{4} leq 1 $



$ -4 leq x + 1 leq 4 $



$ -5 leq x leq 3 $



Assumption 3:



$ 2 + arccosfrac{x+1}{4} neq 0 $



Here once again, no idea how to proceed further.



Would anyone be able to help me understand these examples, how to find that the denominator is not equal to $0$, when there's arccos or arcsin in the denominator? Thanks! I do hope my tags are okay...







functions trigonometry






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weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 days ago









N. F. Taussig

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asked 2 days ago









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weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






weno is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • Use en.wikipedia.org/wiki/…
    – lab bhattacharjee
    2 days ago


















  • Use en.wikipedia.org/wiki/…
    – lab bhattacharjee
    2 days ago
















Use en.wikipedia.org/wiki/…
– lab bhattacharjee
2 days ago




Use en.wikipedia.org/wiki/…
– lab bhattacharjee
2 days ago










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For the first function, the fraction is defined when $arcsin frac{x}{4} ne 0 implies frac{x}{4} ne sin 0=0 implies xne 0$



So the domain of the first function is $[-4,4]-{0}$



For the second function $$2+arccos frac{x+1}{4} ne 0$$
$$implies arccos frac{x+1}{4} ne -2$$
Apply cos both sides,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$






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  • 2




    But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
    – Lubin
    2 days ago










  • The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
    – N. F. Taussig
    yesterday






  • 1




    Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
    – N. F. Taussig
    yesterday











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For the first function, the fraction is defined when $arcsin frac{x}{4} ne 0 implies frac{x}{4} ne sin 0=0 implies xne 0$



So the domain of the first function is $[-4,4]-{0}$



For the second function $$2+arccos frac{x+1}{4} ne 0$$
$$implies arccos frac{x+1}{4} ne -2$$
Apply cos both sides,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$






share|cite|improve this answer



















  • 2




    But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
    – Lubin
    2 days ago










  • The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
    – N. F. Taussig
    yesterday






  • 1




    Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
    – N. F. Taussig
    yesterday















up vote
0
down vote













For the first function, the fraction is defined when $arcsin frac{x}{4} ne 0 implies frac{x}{4} ne sin 0=0 implies xne 0$



So the domain of the first function is $[-4,4]-{0}$



For the second function $$2+arccos frac{x+1}{4} ne 0$$
$$implies arccos frac{x+1}{4} ne -2$$
Apply cos both sides,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$






share|cite|improve this answer



















  • 2




    But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
    – Lubin
    2 days ago










  • The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
    – N. F. Taussig
    yesterday






  • 1




    Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
    – N. F. Taussig
    yesterday













up vote
0
down vote










up vote
0
down vote









For the first function, the fraction is defined when $arcsin frac{x}{4} ne 0 implies frac{x}{4} ne sin 0=0 implies xne 0$



So the domain of the first function is $[-4,4]-{0}$



For the second function $$2+arccos frac{x+1}{4} ne 0$$
$$implies arccos frac{x+1}{4} ne -2$$
Apply cos both sides,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$






share|cite|improve this answer














For the first function, the fraction is defined when $arcsin frac{x}{4} ne 0 implies frac{x}{4} ne sin 0=0 implies xne 0$



So the domain of the first function is $[-4,4]-{0}$



For the second function $$2+arccos frac{x+1}{4} ne 0$$
$$implies arccos frac{x+1}{4} ne -2$$
Apply cos both sides,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 23 hours ago

























answered 2 days ago









Fareed AF

31411




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  • 2




    But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
    – Lubin
    2 days ago










  • The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
    – N. F. Taussig
    yesterday






  • 1




    Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
    – N. F. Taussig
    yesterday














  • 2




    But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
    – Lubin
    2 days ago










  • The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
    – N. F. Taussig
    yesterday






  • 1




    Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
    – N. F. Taussig
    yesterday








2




2




But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
– Lubin
2 days ago




But since $arcsin$ is defined only on the closed interval $[-1,1]$, the denominator in #1 is defined only on the closed interval $[-4,4]$. Outside of that interval, the whole is undefined.
– Lubin
2 days ago












The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
– N. F. Taussig
yesterday




The range of $arccos x$ is $[0, pi]$. Hence, the range of $2 + arccosleft(frac{x + 1}{4}right)$ is $[2, 2 + pi]$.
– N. F. Taussig
yesterday




1




1




Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
– N. F. Taussig
yesterday




Type $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, $arctan x$ to produce $sin x$, $cos x$, $tan x$, $csc x$, $sec x$, $cot x$, $arcsin x$, $arccos x$, and $arctan x$, respectively.
– N. F. Taussig
yesterday










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