Combining functions and finding domain
up vote
1
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favorite
$f(x) = ln (2 - x)$
$g(x) = frac {1}{√x}$
I am asked to find the domain of $frac gf$
So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$
so the domain is $0 < x < 2$
Is this correct?
Thank you!
algebra-precalculus
edited Nov 22 at 18:32
gimusi
92.7k94495
asked Nov 22 at 18:04
M Do
124
add a comment |
up vote
1
down vote
favorite
$f(x) = ln (2 - x)$
$g(x) = frac {1}{√x}$
I am asked to find the domain of $frac gf$
So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$
so the domain is $0 < x < 2$
Is this correct?
Thank you!
algebra-precalculus
edited Nov 22 at 18:32
gimusi
92.7k94495
asked Nov 22 at 18:04
M Do
124
$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06
Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09
No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
$f(x) = ln (2 - x)$
$g(x) = frac {1}{√x}$
I am asked to find the domain of $frac gf$
So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$
so the domain is $0 < x < 2$
Is this correct?
Thank you!
algebra-precalculus
edited Nov 22 at 18:32
gimusi
92.7k94495
asked Nov 22 at 18:04
M Do
124
$f(x) = ln (2 - x)$
$g(x) = frac {1}{√x}$
I am asked to find the domain of $frac gf$
So $frac gf$ would be $frac {1}{√x}$ over $ln (2 - x)
$
which would be $frac{ln (2-x)}{√x}$
so the domain is $0 < x < 2$
Is this correct?
Thank you!
algebra-precalculus
algebra-precalculus
edited Nov 22 at 18:32
gimusi
92.7k94495
asked Nov 22 at 18:04
M Do
124
edited Nov 22 at 18:32
gimusi
92.7k94495
asked Nov 22 at 18:04
M Do
124
edited Nov 22 at 18:32
gimusi
92.7k94495
edited Nov 22 at 18:32
gimusi
92.7k94495
edited Nov 22 at 18:32
gimusi
92.7k94495
92.7k94495
asked Nov 22 at 18:04
M Do
124
asked Nov 22 at 18:04
M Do
124
asked Nov 22 at 18:04
M Do
124
124
$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06
Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09
No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10
add a comment |
$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06
Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09
No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10
$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06
$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06
Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09
Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09
No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10
No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10
add a comment |
1 Answer
1
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oldest
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up vote
0
down vote
HINT
Note that
$$frac g f = frac1{sqrt x log (2-x)}$$
and we need
- for the definition of $f$: $$2-x>0$$
- for the definition of $g$: $$xge 0$$
- for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$
for there we obtain that
$x>0$- $2-x>0$
- $2-xneq 1$
Solve that system to find the domain.
answered Nov 22 at 18:10
gimusi
92.7k94495
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
up vote
0
down vote
HINT
Note that
$$frac g f = frac1{sqrt x log (2-x)}$$
and we need
- for the definition of $f$: $$2-x>0$$
- for the definition of $g$: $$xge 0$$
- for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$
for there we obtain that
$x>0$- $2-x>0$
- $2-xneq 1$
Solve that system to find the domain.
answered Nov 22 at 18:10
gimusi
92.7k94495
add a comment |
up vote
0
down vote
HINT
Note that
$$frac g f = frac1{sqrt x log (2-x)}$$
and we need
- for the definition of $f$: $$2-x>0$$
- for the definition of $g$: $$xge 0$$
- for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$
for there we obtain that
$x>0$- $2-x>0$
- $2-xneq 1$
Solve that system to find the domain.
answered Nov 22 at 18:10
gimusi
92.7k94495
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT
Note that
$$frac g f = frac1{sqrt x log (2-x)}$$
and we need
- for the definition of $f$: $$2-x>0$$
- for the definition of $g$: $$xge 0$$
- for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$
for there we obtain that
$x>0$- $2-x>0$
- $2-xneq 1$
Solve that system to find the domain.
answered Nov 22 at 18:10
gimusi
92.7k94495
HINT
Note that
$$frac g f = frac1{sqrt x log (2-x)}$$
and we need
- for the definition of $f$: $$2-x>0$$
- for the definition of $g$: $$xge 0$$
- for the definition of $f/g$: $$sqrt x log (2-x)neq 0$$
for there we obtain that
$x>0$- $2-x>0$
- $2-xneq 1$
Solve that system to find the domain.
answered Nov 22 at 18:10
gimusi
92.7k94495
edited Nov 22 at 18:31
answered Nov 22 at 18:10
gimusi
92.7k94495
answered Nov 22 at 18:10
gimusi
92.7k94495
answered Nov 22 at 18:10
gimusi
92.7k94495
92.7k94495
add a comment |
add a comment |
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$g(x)/f(x)$ is not the function you wrote. Check it.
– james watt
Nov 22 at 18:06
Edited. I think my domain is still right?
– M Do
Nov 22 at 18:09
No, $g/f$ is not correct.
– james watt
Nov 22 at 18:10