Mixing
If $f(x) = dfrac{x - 2011}{11}$, then evaluate $(f circ f circ f circ f circ f)(x)$
$begingroup$
If $f(x) = dfrac{x - 2011}{11}$, then $(f circ f circ f circ f circ f)(x)$ is $cdots$
A. $frac{x+2011}{x-1}$
B. $frac{x+2011}{x+1}$
C. $frac{x-2011}{x+1}$
D. $frac{x-2011}{x-1}$
E. $frac{-x+2011}{x-1}$
I find no leading to solve this one. Please help me.
I thought there is a sort of pattern so that we can find the faster way for it, but it looks like the first composition already makes the thing too complicated.
Can you give me some tips?
functions function-and-relation-composition
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
$endgroup$
add a comment |
$begingroup$
If $f(x) = dfrac{x - 2011}{11}$, then $(f circ f circ f circ f circ f)(x)$ is $cdots$
A. $frac{x+2011}{x-1}$
B. $frac{x+2011}{x+1}$
C. $frac{x-2011}{x+1}$
D. $frac{x-2011}{x-1}$
E. $frac{-x+2011}{x-1}$
I find no leading to solve this one. Please help me.
I thought there is a sort of pattern so that we can find the faster way for it, but it looks like the first composition already makes the thing too complicated.
Can you give me some tips?
functions function-and-relation-composition
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
$endgroup$
3
$begingroup$
None of them You must have made a mistake in copying the question.
$endgroup$
– TonyK
Jan 17 at 15:21
$begingroup$
Indeed, it should be a linear function.
$endgroup$
– Harnak
Jan 17 at 15:25
$begingroup$
Looks like the right answer is just a line with big numbers: $(x-32387155)/161051$.
$endgroup$
– Dan Uznanski
Jan 17 at 15:26
add a comment |
$begingroup$
If $f(x) = dfrac{x - 2011}{11}$, then $(f circ f circ f circ f circ f)(x)$ is $cdots$
A. $frac{x+2011}{x-1}$
B. $frac{x+2011}{x+1}$
C. $frac{x-2011}{x+1}$
D. $frac{x-2011}{x-1}$
E. $frac{-x+2011}{x-1}$
I find no leading to solve this one. Please help me.
I thought there is a sort of pattern so that we can find the faster way for it, but it looks like the first composition already makes the thing too complicated.
Can you give me some tips?
functions function-and-relation-composition
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
$endgroup$
If $f(x) = dfrac{x - 2011}{11}$, then $(f circ f circ f circ f circ f)(x)$ is $cdots$
A. $frac{x+2011}{x-1}$
B. $frac{x+2011}{x+1}$
C. $frac{x-2011}{x+1}$
D. $frac{x-2011}{x-1}$
E. $frac{-x+2011}{x-1}$
I find no leading to solve this one. Please help me.
I thought there is a sort of pattern so that we can find the faster way for it, but it looks like the first composition already makes the thing too complicated.
Can you give me some tips?
functions function-and-relation-composition
functions function-and-relation-composition
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
asked Jan 17 at 15:17
Shane Dizzy SukardyShane Dizzy Sukardy
60818
60818
3
$begingroup$
None of them You must have made a mistake in copying the question.
$endgroup$
– TonyK
Jan 17 at 15:21
$begingroup$
Indeed, it should be a linear function.
$endgroup$
– Harnak
Jan 17 at 15:25
$begingroup$
Looks like the right answer is just a line with big numbers: $(x-32387155)/161051$.
$endgroup$
– Dan Uznanski
Jan 17 at 15:26
add a comment |
3
$begingroup$
None of them You must have made a mistake in copying the question.
$endgroup$
– TonyK
Jan 17 at 15:21
$begingroup$
Indeed, it should be a linear function.
$endgroup$
– Harnak
Jan 17 at 15:25
$begingroup$
Looks like the right answer is just a line with big numbers: $(x-32387155)/161051$.
$endgroup$
– Dan Uznanski
Jan 17 at 15:26
3
3
$begingroup$
None of them You must have made a mistake in copying the question.
$endgroup$
– TonyK
Jan 17 at 15:21
$begingroup$
None of them You must have made a mistake in copying the question.
$endgroup$
– TonyK
Jan 17 at 15:21
$begingroup$
Indeed, it should be a linear function.
$endgroup$
– Harnak
Jan 17 at 15:25
$begingroup$
Indeed, it should be a linear function.
$endgroup$
– Harnak
Jan 17 at 15:25
$begingroup$
Looks like the right answer is just a line with big numbers: $(x-32387155)/161051$.
$endgroup$
– Dan Uznanski
Jan 17 at 15:26
$begingroup$
Looks like the right answer is just a line with big numbers: $(x-32387155)/161051$.
$endgroup$
– Dan Uznanski
Jan 17 at 15:26
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: The given answers are incorrect. As for solving the problem, why not try a general case and see if any patterns emerge?
$$f(x) = frac{xcolor{blue}{+b}}{color{purple}{c}}$$
$$(fcirc f)(x) = frac{frac{x+b}{c}+b}{c} = frac{frac{x+b+bc}{c}}{c} = frac{xcolor{blue}{+b+bc}}{color{purple}{c^2}}$$
$$(fcirc fcirc f)(x) = frac{frac{x+b+bc}{c^2}+b}{c} = frac{frac{x+b+bc+bc^2}{c^2}}{c} = frac{xcolor{blue}{+b+bc+bc^2}}{color{purple}{c^3}}$$
Notice the geometric progression in the numerators.
answered Jan 17 at 15:27
KM101KM101
6,0251525
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: The given answers are incorrect. As for solving the problem, why not try a general case and see if any patterns emerge?
$$f(x) = frac{xcolor{blue}{+b}}{color{purple}{c}}$$
$$(fcirc f)(x) = frac{frac{x+b}{c}+b}{c} = frac{frac{x+b+bc}{c}}{c} = frac{xcolor{blue}{+b+bc}}{color{purple}{c^2}}$$
$$(fcirc fcirc f)(x) = frac{frac{x+b+bc}{c^2}+b}{c} = frac{frac{x+b+bc+bc^2}{c^2}}{c} = frac{xcolor{blue}{+b+bc+bc^2}}{color{purple}{c^3}}$$
Notice the geometric progression in the numerators.
answered Jan 17 at 15:27
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
Hint: The given answers are incorrect. As for solving the problem, why not try a general case and see if any patterns emerge?
$$f(x) = frac{xcolor{blue}{+b}}{color{purple}{c}}$$
$$(fcirc f)(x) = frac{frac{x+b}{c}+b}{c} = frac{frac{x+b+bc}{c}}{c} = frac{xcolor{blue}{+b+bc}}{color{purple}{c^2}}$$
$$(fcirc fcirc f)(x) = frac{frac{x+b+bc}{c^2}+b}{c} = frac{frac{x+b+bc+bc^2}{c^2}}{c} = frac{xcolor{blue}{+b+bc+bc^2}}{color{purple}{c^3}}$$
Notice the geometric progression in the numerators.
answered Jan 17 at 15:27
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
Hint: The given answers are incorrect. As for solving the problem, why not try a general case and see if any patterns emerge?
$$f(x) = frac{xcolor{blue}{+b}}{color{purple}{c}}$$
$$(fcirc f)(x) = frac{frac{x+b}{c}+b}{c} = frac{frac{x+b+bc}{c}}{c} = frac{xcolor{blue}{+b+bc}}{color{purple}{c^2}}$$
$$(fcirc fcirc f)(x) = frac{frac{x+b+bc}{c^2}+b}{c} = frac{frac{x+b+bc+bc^2}{c^2}}{c} = frac{xcolor{blue}{+b+bc+bc^2}}{color{purple}{c^3}}$$
Notice the geometric progression in the numerators.
answered Jan 17 at 15:27
KM101KM101
6,0251525
$endgroup$
Hint: The given answers are incorrect. As for solving the problem, why not try a general case and see if any patterns emerge?
$$f(x) = frac{xcolor{blue}{+b}}{color{purple}{c}}$$
$$(fcirc f)(x) = frac{frac{x+b}{c}+b}{c} = frac{frac{x+b+bc}{c}}{c} = frac{xcolor{blue}{+b+bc}}{color{purple}{c^2}}$$
$$(fcirc fcirc f)(x) = frac{frac{x+b+bc}{c^2}+b}{c} = frac{frac{x+b+bc+bc^2}{c^2}}{c} = frac{xcolor{blue}{+b+bc+bc^2}}{color{purple}{c^3}}$$
Notice the geometric progression in the numerators.
answered Jan 17 at 15:27
KM101KM101
6,0251525
answered Jan 17 at 15:27
KM101KM101
6,0251525
answered Jan 17 at 15:27
KM101KM101
6,0251525
answered Jan 17 at 15:27
KM101KM101
6,0251525
6,0251525
add a comment |
add a comment |
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3
$begingroup$
None of them You must have made a mistake in copying the question.
$endgroup$
– TonyK
Jan 17 at 15:21
$begingroup$
Indeed, it should be a linear function.
$endgroup$
– Harnak
Jan 17 at 15:25
$begingroup$
Looks like the right answer is just a line with big numbers: $(x-32387155)/161051$.
$endgroup$
– Dan Uznanski
Jan 17 at 15:26