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...
functions trigonometry
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add a comment |
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...
functions trigonometry
New contributor
Use en.wikipedia.org/wiki/…
– lab bhattacharjee
2 days ago
add a comment |
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...
functions trigonometry
New contributor
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
functions trigonometry
New contributor
New contributor
edited 2 days ago
N. F. Taussig
42.6k93254
42.6k93254
New contributor
asked 2 days ago
weno
303
303
New contributor
New contributor
Use en.wikipedia.org/wiki/…
– lab bhattacharjee
2 days ago
add a comment |
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
add a comment |
1 Answer
1
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oldest
votes
up vote
0
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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,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$
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
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$
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
add a comment |
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,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$
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
add a comment |
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,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$
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,
$$implies frac{x+1}{4} ne cos(-2)$$
$$implies x ne 4cos(-2) -1$$
edited 23 hours ago
answered 2 days ago
Fareed AF
31411
31411
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
add a comment |
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
add a comment |
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Use en.wikipedia.org/wiki/…
– lab bhattacharjee
2 days ago