Interval Notation for Increasing and Decreasing Intervals of a Function
1
$begingroup$
This was brought up by another student in one of my pre-calculus classes.
The graph was a simple quadratic $x^2$. The teacher stated that the graph was decreasing from $(-infty,0)$, and increasing from $(0, infty)$.
Why would zero not be included? i.e: decr. $(-infty,0]$ and incr. $[0, infty)$
functions notation quadratics
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
$endgroup$
add a comment |
1
$begingroup$
This was brought up by another student in one of my pre-calculus classes.
The graph was a simple quadratic $x^2$. The teacher stated that the graph was decreasing from $(-infty,0)$, and increasing from $(0, infty)$.
Why would zero not be included? i.e: decr. $(-infty,0]$ and incr. $[0, infty)$
functions notation quadratics
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
$endgroup$
add a comment |
1
1
1
1
$begingroup$
This was brought up by another student in one of my pre-calculus classes.
The graph was a simple quadratic $x^2$. The teacher stated that the graph was decreasing from $(-infty,0)$, and increasing from $(0, infty)$.
Why would zero not be included? i.e: decr. $(-infty,0]$ and incr. $[0, infty)$
functions notation quadratics
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
$endgroup$
This was brought up by another student in one of my pre-calculus classes.
The graph was a simple quadratic $x^2$. The teacher stated that the graph was decreasing from $(-infty,0)$, and increasing from $(0, infty)$.
Why would zero not be included? i.e: decr. $(-infty,0]$ and incr. $[0, infty)$
functions notation quadratics
functions notation quadratics
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
asked Sep 20 '16 at 23:51
AnonymousAnonymous
134
134
add a comment |
add a comment |
1 Answer
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Generally the $0$ is not included because the function is not decreasing (or increasing) at $0$.
It would be accurate, however to say that $y=x^2$ is non-increasing on the interval $(-infty,0]$.
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
$endgroup$
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
add a comment |
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1 Answer
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1 Answer
1
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0
$begingroup$
Generally the $0$ is not included because the function is not decreasing (or increasing) at $0$.
It would be accurate, however to say that $y=x^2$ is non-increasing on the interval $(-infty,0]$.
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
$endgroup$
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
add a comment |
0
$begingroup$
Generally the $0$ is not included because the function is not decreasing (or increasing) at $0$.
It would be accurate, however to say that $y=x^2$ is non-increasing on the interval $(-infty,0]$.
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
$endgroup$
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
add a comment |
0
0
0
$begingroup$
Generally the $0$ is not included because the function is not decreasing (or increasing) at $0$.
It would be accurate, however to say that $y=x^2$ is non-increasing on the interval $(-infty,0]$.
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
$endgroup$
Generally the $0$ is not included because the function is not decreasing (or increasing) at $0$.
It would be accurate, however to say that $y=x^2$ is non-increasing on the interval $(-infty,0]$.
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
answered Sep 21 '16 at 0:03
John Wayland BalesJohn Wayland Bales
14.2k21238
14.2k21238
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
add a comment |
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
But given any single point on any function it will not be increasing or decreasing. It is only increasing/decreasing relative to the points surrounding it. If I were to take the points (-1, 1) and (0,0), as I am going right, it would be decreasing, correct? So is it just a preference or is there mathematical reason behind it?
$endgroup$
– Anonymous
Sep 21 '16 at 0:08
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
$begingroup$
The difficulty arises from the fact that two different definitions are generally given for what it means for a function to decrease (or increase) on an interval $I$. One definition says that $f$ decreases on $I$ if for $a<b$ in $I$ it is true that $f(a)>f(b)$. According to this definition, $f(x)=x^2$ is decreasing on the interval $(-infty,0]$. Another definition says that $f$ decreases on $I$ if for $xin I$, $f^prime(x)<0.$ According to this definition, $f(x)=x^2$ is not decreasing on the interval $(-infty,0]$.So the 'correct' answer depends upon which definition is being used.
$endgroup$
– John Wayland Bales
Sep 21 '16 at 21:48
add a comment |
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