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Neat trick to see when a function is positive and when it is negative and finding inflection points.
Clash Royale CLAN TAG#URR8PPP
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So I am going through problems about derivatives, inflection points etc. Now When I am given a differentiable function, and I am supposed to find the interval when it increases/decreases. I found this neat trick to find that out. What I do is find all the zeroes of the derivative, then by Darboux's theorem between consecutive zeroes the function has to be either all positive or all negative, is that right? This saves me a lot of time since I do not have to draw the function itself. My question here, is this correct and does it always work, perhaps there are easier ways?
Also, regarding inflection points. By definition, it is a point, where the concavity changes, so where the second derivative changes the sign. Clearly, the necessary condition is that the inflection point is the critical point of $f'(x)$ but what are some other nice conditions that do not require me to know if the sign actually changes (to avoid drawing it). I think I have read that is is necessary but still not sufficient that all the derivatives have to equal to 0 at the point for it to be an inflection point.
I am trying to raise my proficiency/speed due to the nature of the GRE. Thank you!
derivatives
asked Jul 29 at 10:29
Sorfosh
910616
add a comment |Â
up vote
0
down vote
favorite
So I am going through problems about derivatives, inflection points etc. Now When I am given a differentiable function, and I am supposed to find the interval when it increases/decreases. I found this neat trick to find that out. What I do is find all the zeroes of the derivative, then by Darboux's theorem between consecutive zeroes the function has to be either all positive or all negative, is that right? This saves me a lot of time since I do not have to draw the function itself. My question here, is this correct and does it always work, perhaps there are easier ways?
Also, regarding inflection points. By definition, it is a point, where the concavity changes, so where the second derivative changes the sign. Clearly, the necessary condition is that the inflection point is the critical point of $f'(x)$ but what are some other nice conditions that do not require me to know if the sign actually changes (to avoid drawing it). I think I have read that is is necessary but still not sufficient that all the derivatives have to equal to 0 at the point for it to be an inflection point.
I am trying to raise my proficiency/speed due to the nature of the GRE. Thank you!
derivatives
asked Jul 29 at 10:29
Sorfosh
910616
In an interval where a function has no real root and is continous , you can be sure that it is either always positive or always negative in this interval. The intervals where a function decreases / increases can sometimes be also detected with the first derivate.
– Peter
Jul 29 at 12:41
For an inflection point, the necessary condition is that the SECOND derivate is zero. Sufficient is that the THIRD derivate is non-zero and also that the second derivate changes its sign at the determined position. In practice, often the third derivate is non-zero allowing to prove the inflection point immediately.
– Peter
Jul 29 at 12:45
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
So I am going through problems about derivatives, inflection points etc. Now When I am given a differentiable function, and I am supposed to find the interval when it increases/decreases. I found this neat trick to find that out. What I do is find all the zeroes of the derivative, then by Darboux's theorem between consecutive zeroes the function has to be either all positive or all negative, is that right? This saves me a lot of time since I do not have to draw the function itself. My question here, is this correct and does it always work, perhaps there are easier ways?
Also, regarding inflection points. By definition, it is a point, where the concavity changes, so where the second derivative changes the sign. Clearly, the necessary condition is that the inflection point is the critical point of $f'(x)$ but what are some other nice conditions that do not require me to know if the sign actually changes (to avoid drawing it). I think I have read that is is necessary but still not sufficient that all the derivatives have to equal to 0 at the point for it to be an inflection point.
I am trying to raise my proficiency/speed due to the nature of the GRE. Thank you!
derivatives
asked Jul 29 at 10:29
Sorfosh
910616
So I am going through problems about derivatives, inflection points etc. Now When I am given a differentiable function, and I am supposed to find the interval when it increases/decreases. I found this neat trick to find that out. What I do is find all the zeroes of the derivative, then by Darboux's theorem between consecutive zeroes the function has to be either all positive or all negative, is that right? This saves me a lot of time since I do not have to draw the function itself. My question here, is this correct and does it always work, perhaps there are easier ways?
Also, regarding inflection points. By definition, it is a point, where the concavity changes, so where the second derivative changes the sign. Clearly, the necessary condition is that the inflection point is the critical point of $f'(x)$ but what are some other nice conditions that do not require me to know if the sign actually changes (to avoid drawing it). I think I have read that is is necessary but still not sufficient that all the derivatives have to equal to 0 at the point for it to be an inflection point.
I am trying to raise my proficiency/speed due to the nature of the GRE. Thank you!
derivatives
asked Jul 29 at 10:29
Sorfosh
910616
asked Jul 29 at 10:29
Sorfosh
910616
asked Jul 29 at 10:29
Sorfosh
910616
asked Jul 29 at 10:29
Sorfosh
910616
910616
In an interval where a function has no real root and is continous , you can be sure that it is either always positive or always negative in this interval. The intervals where a function decreases / increases can sometimes be also detected with the first derivate.
– Peter
Jul 29 at 12:41
For an inflection point, the necessary condition is that the SECOND derivate is zero. Sufficient is that the THIRD derivate is non-zero and also that the second derivate changes its sign at the determined position. In practice, often the third derivate is non-zero allowing to prove the inflection point immediately.
– Peter
Jul 29 at 12:45
add a comment |Â
In an interval where a function has no real root and is continous , you can be sure that it is either always positive or always negative in this interval. The intervals where a function decreases / increases can sometimes be also detected with the first derivate.
– Peter
Jul 29 at 12:41
For an inflection point, the necessary condition is that the SECOND derivate is zero. Sufficient is that the THIRD derivate is non-zero and also that the second derivate changes its sign at the determined position. In practice, often the third derivate is non-zero allowing to prove the inflection point immediately.
– Peter
Jul 29 at 12:45
In an interval where a function has no real root and is continous , you can be sure that it is either always positive or always negative in this interval. The intervals where a function decreases / increases can sometimes be also detected with the first derivate.
– Peter
Jul 29 at 12:41
In an interval where a function has no real root and is continous , you can be sure that it is either always positive or always negative in this interval. The intervals where a function decreases / increases can sometimes be also detected with the first derivate.
– Peter
Jul 29 at 12:41
For an inflection point, the necessary condition is that the SECOND derivate is zero. Sufficient is that the THIRD derivate is non-zero and also that the second derivate changes its sign at the determined position. In practice, often the third derivate is non-zero allowing to prove the inflection point immediately.
– Peter
Jul 29 at 12:45
For an inflection point, the necessary condition is that the SECOND derivate is zero. Sufficient is that the THIRD derivate is non-zero and also that the second derivate changes its sign at the determined position. In practice, often the third derivate is non-zero allowing to prove the inflection point immediately.
– Peter
Jul 29 at 12:45
add a comment |Â
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In an interval where a function has no real root and is continous , you can be sure that it is either always positive or always negative in this interval. The intervals where a function decreases / increases can sometimes be also detected with the first derivate.
– Peter
Jul 29 at 12:41
For an inflection point, the necessary condition is that the SECOND derivate is zero. Sufficient is that the THIRD derivate is non-zero and also that the second derivate changes its sign at the determined position. In practice, often the third derivate is non-zero allowing to prove the inflection point immediately.
– Peter
Jul 29 at 12:45