Graphin sin (e^x)
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Hi I was wondering if the limits of x were [0 < x < 360] how would you be able to graph f(x) = sin (e^x). I only was able to workout the range of the graph which will be -1 < f(x) < 1 but not sure how to draw the graph.
Is it possible to obtain a more "friendlier" version so we can apply transformation rule to the function?
graphing-functions
asked Dec 24 '18 at 14:31
john smithjohn smith
133
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add a comment |
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Hi I was wondering if the limits of x were [0 < x < 360] how would you be able to graph f(x) = sin (e^x). I only was able to workout the range of the graph which will be -1 < f(x) < 1 but not sure how to draw the graph.
Is it possible to obtain a more "friendlier" version so we can apply transformation rule to the function?
graphing-functions
asked Dec 24 '18 at 14:31
john smithjohn smith
133
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$begingroup$
What do you mean by a friendlier version?
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– Ross Millikan
Dec 24 '18 at 14:47
add a comment |
$begingroup$
Hi I was wondering if the limits of x were [0 < x < 360] how would you be able to graph f(x) = sin (e^x). I only was able to workout the range of the graph which will be -1 < f(x) < 1 but not sure how to draw the graph.
Is it possible to obtain a more "friendlier" version so we can apply transformation rule to the function?
graphing-functions
asked Dec 24 '18 at 14:31
john smithjohn smith
133
$endgroup$
Hi I was wondering if the limits of x were [0 < x < 360] how would you be able to graph f(x) = sin (e^x). I only was able to workout the range of the graph which will be -1 < f(x) < 1 but not sure how to draw the graph.
Is it possible to obtain a more "friendlier" version so we can apply transformation rule to the function?
graphing-functions
graphing-functions
asked Dec 24 '18 at 14:31
john smithjohn smith
133
asked Dec 24 '18 at 14:31
john smithjohn smith
133
asked Dec 24 '18 at 14:31
john smithjohn smith
133
asked Dec 24 '18 at 14:31
john smithjohn smith
133
asked Dec 24 '18 at 14:31
john smithjohn smith
133
133
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What do you mean by a friendlier version?
$endgroup$
– Ross Millikan
Dec 24 '18 at 14:47
add a comment |
$begingroup$
What do you mean by a friendlier version?
$endgroup$
– Ross Millikan
Dec 24 '18 at 14:47
$begingroup$
What do you mean by a friendlier version?
$endgroup$
– Ross Millikan
Dec 24 '18 at 14:47
$begingroup$
What do you mean by a friendlier version?
$endgroup$
– Ross Millikan
Dec 24 '18 at 14:47
add a comment |
1 Answer
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Since $e^{360}approx 2.2cdot 10^{156}$ is so huge, $sin(e^x)$ will go over an enormous number of cycles. The graph will just look like a solid band once $x$ gets larger than a few. You won't be able to see the details of the graph at all.
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
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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
$begingroup$
Since $e^{360}approx 2.2cdot 10^{156}$ is so huge, $sin(e^x)$ will go over an enormous number of cycles. The graph will just look like a solid band once $x$ gets larger than a few. You won't be able to see the details of the graph at all.
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Since $e^{360}approx 2.2cdot 10^{156}$ is so huge, $sin(e^x)$ will go over an enormous number of cycles. The graph will just look like a solid band once $x$ gets larger than a few. You won't be able to see the details of the graph at all.
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
$endgroup$
add a comment |
$begingroup$
Since $e^{360}approx 2.2cdot 10^{156}$ is so huge, $sin(e^x)$ will go over an enormous number of cycles. The graph will just look like a solid band once $x$ gets larger than a few. You won't be able to see the details of the graph at all.
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
$endgroup$
Since $e^{360}approx 2.2cdot 10^{156}$ is so huge, $sin(e^x)$ will go over an enormous number of cycles. The graph will just look like a solid band once $x$ gets larger than a few. You won't be able to see the details of the graph at all.
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
answered Dec 24 '18 at 14:46
Ross MillikanRoss Millikan
301k24200375
301k24200375
add a comment |
add a comment |
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What do you mean by a friendlier version?
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– Ross Millikan
Dec 24 '18 at 14:47