View Full Version : Stationary Points of Inflection
Cummings
May31-04, 05:31 AM
Now, given y=x^3 -9x^2+23x-16 on the interval [-3,7]
the maximum and minimum values would be the turning points right?
also, a stationary point of inflextion is where the grandient is zero, with a positive or negative gradient on both sides right?
i am asked to find the EXACT values of the x-coordinates of the points of inflection on the graph of -x^4 + 3x^3 + 5x^2 + 2x + 11
but, there are two maximums and one minimum, not a stationary point of inflection.
So, are minimum and maximum values points of inflections?
Also, how do i obtain an EXACT value for the x coordinates.
arildno
May31-04, 09:09 AM
Now, given y=x^3 -9x^2+23x-16 on the interval [-3,7]
the maximum and minimum values would be the turning points right?
also, a stationary point of inflextion is where the grandient is zero, with a positive or negative gradient on both sides right?
i am asked to find the EXACT values of the x-coordinates of the points of inflection on the graph of -x^4 + 3x^3 + 5x^2 + 2x + 11
but, there are two maximums and one minimum, not a stationary point of inflection.
So, are minimum and maximum values points of inflections?
Also, how do i obtain an EXACT value for the x coordinates.
You are mixing together the concepts "point of inflection" and "saddle point"
A "saddle point" is a point where the derivative is zero, but where the function does not achieve a local extremal value.
A "point of inflection" is a point where the curvature changes sign; in particular, the 2.derivative is zero at a "point of inflection".
e(ho0n3
May31-04, 11:38 AM
Now, given y=x^3 -9x^2+23x-16 on the interval [-3,7]
the maximum and minimum values would be the turning points right?
Since you're given an interval, then you're dealing with local extrema here. I'm not sure what you mean by 'turning point'. I suggest you take a look at the definition of a local max. and min. again.
also, a stationary point of inflextion is where the grandient is zero, with a positive or negative gradient on both sides right?
i am asked to find the EXACT values of the x-coordinates of the points of inflection on the graph of -x^4 + 3x^3 + 5x^2 + 2x + 11
but, there are two maximums and one minimum, not a stationary point of inflection.
So, are minimum and maximum values points of inflections?
Also, how do i obtain an EXACT value for the x coordinates.
As arildno said, you're probably getting some of these concepts mixed up. There is a so-called second derivative test which you can use to find whether a given point on the function is a a local maximum or minimum. It basically says:
If for some p, f'(p) = 0 then
- If f''(p) > 0, there is a local minimum at (p, f(p))
- If f''(p) < 0, there is a local maximum at (p, f(p))
- If f''(p) = 0, then you're out of luck.
Hope that helps,
e(ho0n3
confuted
May31-04, 01:44 PM
You are mixing together the concepts "point of inflection" and "saddle point"
A "saddle point" is a point where the derivative is zero, but where the function does not achieve a local extremal value.
A "point of inflection" is a point where the curvature changes sign; in particular, the 2.derivative is zero at a "point of inflection".
Be careful with that. A point of inflection only occurs if the 2nd derivative changes signs, not simply if it is zero. For example, if f(x)=3x, then f''(x)=0 for all values of x, but there are clearly no inflection points.
But if you look at f(x)=x^3, f''(x)=6x and we see that f''(0)=0, f''(-.1)=-.6, f''(.1)=.6 and the 2nd derivative is clearly changing signs at x=0, so there is an inflection point there.
arildno
May31-04, 01:52 PM
Hmm..that's what I meant with ("a point where the curvature changes sign"), but I see now that the last part of the sentence made the meaning ambiguous.
Thx.
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