Mixing
Questions about Functions
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The function $f$ is defined by $f(x)=frac{(m+x)}{(2+3x)}$ for all values of $x$ except when $x=h$ and $m$ is a constant. Given that value of $1$ is unchanged under this mapping, how should I find the value of $m$?
functions
edited Jan 29 at 13:12
John Hughes
65.1k24292
asked Jan 29 at 12:48
LeonLeon
21
$endgroup$
add a comment |
$begingroup$
The function $f$ is defined by $f(x)=frac{(m+x)}{(2+3x)}$ for all values of $x$ except when $x=h$ and $m$ is a constant. Given that value of $1$ is unchanged under this mapping, how should I find the value of $m$?
functions
edited Jan 29 at 13:12
John Hughes
65.1k24292
asked Jan 29 at 12:48
LeonLeon
21
$endgroup$
$begingroup$
@KushalBhuyan: I re-added the dollar signs around "m" and "1" because all mathematics should be set in math-mode. (I also added them around $f$ at the start of the first sentence.) Does that make sense?
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– John Hughes
Jan 29 at 13:14
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@JohnHughes Yes thank you. It does make sense.
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– Kushal Bhuyan
Jan 29 at 14:48
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You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:33
add a comment |
$begingroup$
The function $f$ is defined by $f(x)=frac{(m+x)}{(2+3x)}$ for all values of $x$ except when $x=h$ and $m$ is a constant. Given that value of $1$ is unchanged under this mapping, how should I find the value of $m$?
functions
edited Jan 29 at 13:12
John Hughes
65.1k24292
asked Jan 29 at 12:48
LeonLeon
21
$endgroup$
The function $f$ is defined by $f(x)=frac{(m+x)}{(2+3x)}$ for all values of $x$ except when $x=h$ and $m$ is a constant. Given that value of $1$ is unchanged under this mapping, how should I find the value of $m$?
functions
functions
edited Jan 29 at 13:12
John Hughes
65.1k24292
asked Jan 29 at 12:48
LeonLeon
21
edited Jan 29 at 13:12
John Hughes
65.1k24292
asked Jan 29 at 12:48
LeonLeon
21
edited Jan 29 at 13:12
John Hughes
65.1k24292
edited Jan 29 at 13:12
John Hughes
65.1k24292
edited Jan 29 at 13:12
John Hughes
65.1k24292
65.1k24292
asked Jan 29 at 12:48
LeonLeon
21
asked Jan 29 at 12:48
LeonLeon
21
asked Jan 29 at 12:48
LeonLeon
21
21
$begingroup$
@KushalBhuyan: I re-added the dollar signs around "m" and "1" because all mathematics should be set in math-mode. (I also added them around $f$ at the start of the first sentence.) Does that make sense?
$endgroup$
– John Hughes
Jan 29 at 13:14
$begingroup$
@JohnHughes Yes thank you. It does make sense.
$endgroup$
– Kushal Bhuyan
Jan 29 at 14:48
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:33
add a comment |
$begingroup$
@KushalBhuyan: I re-added the dollar signs around "m" and "1" because all mathematics should be set in math-mode. (I also added them around $f$ at the start of the first sentence.) Does that make sense?
$endgroup$
– John Hughes
Jan 29 at 13:14
$begingroup$
@JohnHughes Yes thank you. It does make sense.
$endgroup$
– Kushal Bhuyan
Jan 29 at 14:48
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:33
$begingroup$
@KushalBhuyan: I re-added the dollar signs around "m" and "1" because all mathematics should be set in math-mode. (I also added them around $f$ at the start of the first sentence.) Does that make sense?
$endgroup$
– John Hughes
Jan 29 at 13:14
$begingroup$
@KushalBhuyan: I re-added the dollar signs around "m" and "1" because all mathematics should be set in math-mode. (I also added them around $f$ at the start of the first sentence.) Does that make sense?
$endgroup$
– John Hughes
Jan 29 at 13:14
$begingroup$
@JohnHughes Yes thank you. It does make sense.
$endgroup$
– Kushal Bhuyan
Jan 29 at 14:48
$begingroup$
@JohnHughes Yes thank you. It does make sense.
$endgroup$
– Kushal Bhuyan
Jan 29 at 14:48
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:33
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint
The sentence
The value of $1$ is unchanged under this mapping
translates into the equation $$f(1)=1.$$
answered Jan 29 at 12:49
5xum5xum
91.9k394161
$endgroup$
add a comment |
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$f(1)=frac{m+1}{2+3}=1 iff m+1=5 iff m=4.$
answered Jan 29 at 12:50
FredFred
48.9k11849
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint
The sentence
The value of $1$ is unchanged under this mapping
translates into the equation $$f(1)=1.$$
answered Jan 29 at 12:49
5xum5xum
91.9k394161
$endgroup$
add a comment |
$begingroup$
Hint
The sentence
The value of $1$ is unchanged under this mapping
translates into the equation $$f(1)=1.$$
answered Jan 29 at 12:49
5xum5xum
91.9k394161
$endgroup$
add a comment |
$begingroup$
Hint
The sentence
The value of $1$ is unchanged under this mapping
translates into the equation $$f(1)=1.$$
answered Jan 29 at 12:49
5xum5xum
91.9k394161
$endgroup$
Hint
The sentence
The value of $1$ is unchanged under this mapping
translates into the equation $$f(1)=1.$$
answered Jan 29 at 12:49
5xum5xum
91.9k394161
answered Jan 29 at 12:49
5xum5xum
91.9k394161
answered Jan 29 at 12:49
5xum5xum
91.9k394161
answered Jan 29 at 12:49
5xum5xum
91.9k394161
91.9k394161
add a comment |
add a comment |
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$f(1)=frac{m+1}{2+3}=1 iff m+1=5 iff m=4.$
answered Jan 29 at 12:50
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
$f(1)=frac{m+1}{2+3}=1 iff m+1=5 iff m=4.$
answered Jan 29 at 12:50
FredFred
48.9k11849
$endgroup$
add a comment |
$begingroup$
$f(1)=frac{m+1}{2+3}=1 iff m+1=5 iff m=4.$
answered Jan 29 at 12:50
FredFred
48.9k11849
$endgroup$
$f(1)=frac{m+1}{2+3}=1 iff m+1=5 iff m=4.$
answered Jan 29 at 12:50
FredFred
48.9k11849
answered Jan 29 at 12:50
FredFred
48.9k11849
answered Jan 29 at 12:50
FredFred
48.9k11849
answered Jan 29 at 12:50
FredFred
48.9k11849
48.9k11849
add a comment |
add a comment |
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$begingroup$
@KushalBhuyan: I re-added the dollar signs around "m" and "1" because all mathematics should be set in math-mode. (I also added them around $f$ at the start of the first sentence.) Does that make sense?
$endgroup$
– John Hughes
Jan 29 at 13:14
$begingroup$
@JohnHughes Yes thank you. It does make sense.
$endgroup$
– Kushal Bhuyan
Jan 29 at 14:48
$begingroup$
You recieved 2 answers to your question. Is any of them what you needed? If so, you should upvote all the useful answers and accept the answer that is most useful to you.
$endgroup$
– 5xum
Feb 11 at 11:33