Transformations on function curve
$begingroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
$endgroup$
add a comment |
$begingroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
$endgroup$
add a comment |
$begingroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
$endgroup$
If we have a function $f$ such that $f:Rto R^+$ where $f(x) = 3^{-x}$, it's graph will be like that
And we have another function $g$ such that $g: Rto R^+$ where $g(x) = f(x+1) = 3^{-x+1} = 3^{1-x}$
And because $g(x) = f(x+1)$ so the curve of $g$ is an image of the curve of $f$ by translation 1 unit to the left
But it turns out that the translation is 1 unit to the right and not to the left
The graph of $f(x+1)$ is supposed to be the same as that of $f(x)$ by translation towards the negative part of x-axis.Why isn't that the case here?
algebra-precalculus functions exponential-function graphing-functions
algebra-precalculus functions exponential-function graphing-functions
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
edited Dec 23 '18 at 23:49
gt6989b
35.2k22557
35.2k22557
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
asked Dec 23 '18 at 23:06
Yousef EssamYousef Essam
335
335
add a comment |
add a comment |
1 Answer
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$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
$endgroup$
You did the arithmetic wrong. If $f(x) = 3^{-x}$ then
$$
g(x) = f(x+1) = 3^{-(x+1)} = 3^{-1-x}.
$$
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
answered Dec 23 '18 at 23:49
gt6989bgt6989b
35.2k22557
35.2k22557
add a comment |
add a comment |
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