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 Nov 19 at 19:51
Bernard
117k637109
asked Nov 19 at 19:11
user546115
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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 Nov 19 at 19:51
Bernard
117k637109
asked Nov 19 at 19:11
user546115
11
add a comment |
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 Nov 19 at 19:51
Bernard
117k637109
asked Nov 19 at 19:11
user546115
11
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 Nov 19 at 19:51
Bernard
117k637109
asked Nov 19 at 19:11
user546115
11
edited Nov 19 at 19:51
Bernard
117k637109
asked Nov 19 at 19:11
user546115
11
edited Nov 19 at 19:51
Bernard
117k637109
edited Nov 19 at 19:51
Bernard
117k637109
edited Nov 19 at 19:51
Bernard
117k637109
117k637109
asked Nov 19 at 19:11
user546115
11
asked Nov 19 at 19:11
user546115
11
asked Nov 19 at 19:11
user546115
11
11
add a comment |
add a comment |
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 Nov 19 at 20:05
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 Nov 19 at 20:05
GrzegorzK
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2 Answers
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2 Answers
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up vote
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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 Nov 19 at 20:05
gimusi
91.7k84495
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 Nov 19 at 20:05
gimusi
91.7k84495
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 Nov 19 at 20:05
gimusi
91.7k84495
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 Nov 19 at 20:05
gimusi
91.7k84495
answered Nov 19 at 20:05
gimusi
91.7k84495
answered Nov 19 at 20:05
gimusi
91.7k84495
answered Nov 19 at 20:05
gimusi
91.7k84495
91.7k84495
add a comment |
add a comment |
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0
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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 Nov 19 at 20:05
GrzegorzK
11
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 Nov 19 at 20:05
GrzegorzK
11
add a comment |
up vote
0
down vote
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 Nov 19 at 20:05
GrzegorzK
11
- 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 Nov 19 at 20:05
GrzegorzK
11
answered Nov 19 at 20:05
GrzegorzK
11
answered Nov 19 at 20:05
GrzegorzK
11
answered Nov 19 at 20:05
GrzegorzK
11
11
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