Mixing
domain of some function
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I need domain the following functions.
Can you help me?
$f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.
calculus
edited 2 days ago
Bernard
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up vote
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I need domain the following functions.
Can you help me?
$f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.
calculus
edited 2 days ago
Bernard
115k637108
asked 2 days ago
user546115
1
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user546115 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
I need domain the following functions.
Can you help me?
$f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.
calculus
edited 2 days ago
Bernard
115k637108
asked 2 days ago
user546115
1
New contributor
user546115 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I need domain the following functions.
Can you help me?
$f(x)= tan x .cot x$ and $g(x)= sqrt {1-sin^2 x}$ and $h(x)= sqrt {(1- sin^2 x)^2}$.
calculus
calculus
edited 2 days ago
Bernard
115k637108
asked 2 days ago
user546115
1
New contributor
user546115 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 days ago
Bernard
115k637108
asked 2 days ago
user546115
1
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edited 2 days ago
Bernard
115k637108
edited 2 days ago
Bernard
115k637108
edited 2 days ago
Bernard
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115k637108
asked 2 days ago
user546115
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asked 2 days ago
user546115
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asked 2 days ago
user546115
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2 Answers
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HINT
$f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$
$g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$
$h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$
answered 2 days ago
gimusi
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- We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.
$h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.
$g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.
answered 2 days ago
GrzegorzK
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT
$f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$
$g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$
$h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$
answered 2 days ago
gimusi
86k74294
add a comment |
up vote
0
down vote
HINT
$f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$
$g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$
$h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$
answered 2 days ago
gimusi
86k74294
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT
$f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$
$g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$
$h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$
answered 2 days ago
gimusi
86k74294
HINT
$f(x)= tan x cdot cot x$ we need $sin xneq 0 , land ,cos x neq 0$
$g(x)= sqrt {1-sin^2 x}$ we need $1-sin^2 xge 0$
$h(x)= sqrt {(1-sin^2 x)^2}$ we need $(1-sin^2 x)^2ge 0$
answered 2 days ago
gimusi
86k74294
answered 2 days ago
gimusi
86k74294
answered 2 days ago
gimusi
86k74294
answered 2 days ago
gimusi
86k74294
86k74294
add a comment |
add a comment |
up vote
0
down vote
- We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.
$h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.
$g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.
answered 2 days ago
GrzegorzK
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- We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.
$h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.
$g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.
answered 2 days ago
GrzegorzK
11
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up vote
0
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up vote
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- We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.
$h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.
$g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.
answered 2 days ago
GrzegorzK
11
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- We have $f(x) = frac{sin(x)}{cos(x)}frac{cos(x)}{sin(x)}$. We have $cos(x) = 0$ if and only if $x in { frac{pi}{2} + kpi mid k in mathbb{Z}}$, and $sin(x) = 0$ if and only if $x in { kpi mid k in mathbb{Z}}$. Hence domain of $f$ is $mathbb{R}setminus { frac{pi}{2} + kpi , kpimid k in mathbb{Z} } = mathbb{R}setminus { frac{kpi}{2} mid k in mathbb{Z} }$.
$h(x)$ is defined as long as $(1-sin^2(x))^2 geq 0$, which always holds. Hence domain of $h$ is $mathbb{R}$.
$g(x)$ is defined as long as $1-sin^2(x) geq 0$. We have for every $x in mathbb{R}$ that $sin(x) in [-1,1]$, hence $sin^2(x) in [0,1]$, hence $1-sin^2(x) in [0,1]$. We thus have that domain of $g$ is $mathbb{R}$.
answered 2 days ago
GrzegorzK
11
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answered 2 days ago
GrzegorzK
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answered 2 days ago
GrzegorzK
11
answered 2 days ago
GrzegorzK
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11
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