How can I decipher which graph belongs to which equation?How can you find the nature of a graph?what kind of equation is this and how can I solve it?Determining the positioning of rational functions without plotting pointsHow to graph a sin & cos waveHow to draw the graph of $|x|+|y|=1+x$ and $y+|y|=x+|x|$.How to find equation for a line on a graphHow to graph $frac(x+3)(x+1)$?How can I write the equation of this graph?How to plot graph onlineHow can I Plot “Change Rate Graph” of Sine Graph
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How can I decipher which graph belongs to which equation?
How can you find the nature of a graph?what kind of equation is this and how can I solve it?Determining the positioning of rational functions without plotting pointsHow to graph a sin & cos waveHow to draw the graph of $|x|+|y|=1+x$ and $y+|y|=x+|x|$.How to find equation for a line on a graphHow to graph $frac(x+3)(x+1)$?How can I write the equation of this graph?How to plot graph onlineHow can I Plot “Change Rate Graph” of Sine Graph
$begingroup$
Aside from plotting points, how else can I tell which graph is which?
algebra-precalculus exponential-function graphing-functions
edited 1 hour ago
Martin Sleziak
45.2k11123278
asked 1 hour ago
user130306user130306
424111
$endgroup$
add a comment |
$begingroup$
Aside from plotting points, how else can I tell which graph is which?
algebra-precalculus exponential-function graphing-functions
edited 1 hour ago
Martin Sleziak
45.2k11123278
asked 1 hour ago
user130306user130306
424111
$endgroup$
$begingroup$
The only apparent differences between the graphs are of the kind "red is greater than blue for negative $x$" or similar. This makes plotting (or at least evaluating) one point a simple and efficient option - and ultimately the only one: If you show any other difference between the functions, you must still argue that this leads to the visible difference between the graohs, i.e., that $f(x)<g(x)$ for some negative $x$ or the like
$endgroup$
– Hagen von Eitzen
1 hour ago
add a comment |
$begingroup$
Aside from plotting points, how else can I tell which graph is which?
algebra-precalculus exponential-function graphing-functions
edited 1 hour ago
Martin Sleziak
45.2k11123278
asked 1 hour ago
user130306user130306
424111
$endgroup$
Aside from plotting points, how else can I tell which graph is which?
algebra-precalculus exponential-function graphing-functions
algebra-precalculus exponential-function graphing-functions
edited 1 hour ago
Martin Sleziak
45.2k11123278
asked 1 hour ago
user130306user130306
424111
edited 1 hour ago
Martin Sleziak
45.2k11123278
asked 1 hour ago
user130306user130306
424111
edited 1 hour ago
Martin Sleziak
45.2k11123278
edited 1 hour ago
Martin Sleziak
45.2k11123278
edited 1 hour ago
Martin Sleziak
45.2k11123278
45.2k11123278
asked 1 hour ago
user130306user130306
424111
asked 1 hour ago
user130306user130306
424111
asked 1 hour ago
user130306user130306
424111
424111
$begingroup$
The only apparent differences between the graphs are of the kind "red is greater than blue for negative $x$" or similar. This makes plotting (or at least evaluating) one point a simple and efficient option - and ultimately the only one: If you show any other difference between the functions, you must still argue that this leads to the visible difference between the graohs, i.e., that $f(x)<g(x)$ for some negative $x$ or the like
$endgroup$
– Hagen von Eitzen
1 hour ago
add a comment |
$begingroup$
The only apparent differences between the graphs are of the kind "red is greater than blue for negative $x$" or similar. This makes plotting (or at least evaluating) one point a simple and efficient option - and ultimately the only one: If you show any other difference between the functions, you must still argue that this leads to the visible difference between the graohs, i.e., that $f(x)<g(x)$ for some negative $x$ or the like
$endgroup$
– Hagen von Eitzen
1 hour ago
$begingroup$
The only apparent differences between the graphs are of the kind "red is greater than blue for negative $x$" or similar. This makes plotting (or at least evaluating) one point a simple and efficient option - and ultimately the only one: If you show any other difference between the functions, you must still argue that this leads to the visible difference between the graohs, i.e., that $f(x)<g(x)$ for some negative $x$ or the like
$endgroup$
– Hagen von Eitzen
1 hour ago
$begingroup$
The only apparent differences between the graphs are of the kind "red is greater than blue for negative $x$" or similar. This makes plotting (or at least evaluating) one point a simple and efficient option - and ultimately the only one: If you show any other difference between the functions, you must still argue that this leads to the visible difference between the graohs, i.e., that $f(x)<g(x)$ for some negative $x$ or the like
$endgroup$
– Hagen von Eitzen
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We know that $ln (2) < ln (7)$, if $x<0$, then
$$xln (2) > x ln (7)$$
$$2^x>7^x$$
Similar for the second case, work with $ln (4)$ and $ln (3)$ to compare the graph.
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
$endgroup$
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
1
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
add a comment |
$begingroup$
Consider the 1st graph.
For x<0, 2x > 7x
So now you need to determine which curve is having higher values for the same values of x
The curve taking higher values for same values of x will always be above the curve taking lower values for same values of x.
So, the red curve is 2x and the blue curve is 7x
Consider the 2nd graph
i) x > 0
3-x > 4-x
So, 3 -x will be higher in the region x<0
ii) x < 0
4-x > 3-x
So, 4-x will be higher in the region x<0
By observing the graph it can be concluded that the red curve is 4-x and the blue curve is 3-x
answered 1 hour ago
Free RadicalFree Radical
18111
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
We know that $ln (2) < ln (7)$, if $x<0$, then
$$xln (2) > x ln (7)$$
$$2^x>7^x$$
Similar for the second case, work with $ln (4)$ and $ln (3)$ to compare the graph.
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
$endgroup$
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
1
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
add a comment |
$begingroup$
We know that $ln (2) < ln (7)$, if $x<0$, then
$$xln (2) > x ln (7)$$
$$2^x>7^x$$
Similar for the second case, work with $ln (4)$ and $ln (3)$ to compare the graph.
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
$endgroup$
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
1
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
add a comment |
$begingroup$
We know that $ln (2) < ln (7)$, if $x<0$, then
$$xln (2) > x ln (7)$$
$$2^x>7^x$$
Similar for the second case, work with $ln (4)$ and $ln (3)$ to compare the graph.
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
$endgroup$
We know that $ln (2) < ln (7)$, if $x<0$, then
$$xln (2) > x ln (7)$$
$$2^x>7^x$$
Similar for the second case, work with $ln (4)$ and $ln (3)$ to compare the graph.
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
answered 1 hour ago
Siong Thye GohSiong Thye Goh
105k1469121
105k1469121
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
1
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
add a comment |
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
1
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
$begingroup$
ah thank you! i have a really dumb question but, why do we know that $ln(2) < ln(7)$? just by nature of the logarithm function or are you just computing it?
$endgroup$
– user130306
1 hour ago
1
1
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
$begingroup$
We know $ln(x)$ is an increasing function.
$endgroup$
– Siong Thye Goh
1 hour ago
add a comment |
$begingroup$
Consider the 1st graph.
For x<0, 2x > 7x
So now you need to determine which curve is having higher values for the same values of x
The curve taking higher values for same values of x will always be above the curve taking lower values for same values of x.
So, the red curve is 2x and the blue curve is 7x
Consider the 2nd graph
i) x > 0
3-x > 4-x
So, 3 -x will be higher in the region x<0
ii) x < 0
4-x > 3-x
So, 4-x will be higher in the region x<0
By observing the graph it can be concluded that the red curve is 4-x and the blue curve is 3-x
answered 1 hour ago
Free RadicalFree Radical
18111
$endgroup$
add a comment |
$begingroup$
Consider the 1st graph.
For x<0, 2x > 7x
So now you need to determine which curve is having higher values for the same values of x
The curve taking higher values for same values of x will always be above the curve taking lower values for same values of x.
So, the red curve is 2x and the blue curve is 7x
Consider the 2nd graph
i) x > 0
3-x > 4-x
So, 3 -x will be higher in the region x<0
ii) x < 0
4-x > 3-x
So, 4-x will be higher in the region x<0
By observing the graph it can be concluded that the red curve is 4-x and the blue curve is 3-x
answered 1 hour ago
Free RadicalFree Radical
18111
$endgroup$
add a comment |
$begingroup$
Consider the 1st graph.
For x<0, 2x > 7x
So now you need to determine which curve is having higher values for the same values of x
The curve taking higher values for same values of x will always be above the curve taking lower values for same values of x.
So, the red curve is 2x and the blue curve is 7x
Consider the 2nd graph
i) x > 0
3-x > 4-x
So, 3 -x will be higher in the region x<0
ii) x < 0
4-x > 3-x
So, 4-x will be higher in the region x<0
By observing the graph it can be concluded that the red curve is 4-x and the blue curve is 3-x
answered 1 hour ago
Free RadicalFree Radical
18111
$endgroup$
Consider the 1st graph.
For x<0, 2x > 7x
So now you need to determine which curve is having higher values for the same values of x
The curve taking higher values for same values of x will always be above the curve taking lower values for same values of x.
So, the red curve is 2x and the blue curve is 7x
Consider the 2nd graph
i) x > 0
3-x > 4-x
So, 3 -x will be higher in the region x<0
ii) x < 0
4-x > 3-x
So, 4-x will be higher in the region x<0
By observing the graph it can be concluded that the red curve is 4-x and the blue curve is 3-x
answered 1 hour ago
Free RadicalFree Radical
18111
answered 1 hour ago
Free RadicalFree Radical
18111
answered 1 hour ago
Free RadicalFree Radical
18111
answered 1 hour ago
Free RadicalFree Radical
18111
18111
add a comment |
add a comment |
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$begingroup$
The only apparent differences between the graphs are of the kind "red is greater than blue for negative $x$" or similar. This makes plotting (or at least evaluating) one point a simple and efficient option - and ultimately the only one: If you show any other difference between the functions, you must still argue that this leads to the visible difference between the graohs, i.e., that $f(x)<g(x)$ for some negative $x$ or the like
$endgroup$
– Hagen von Eitzen
1 hour ago