Sketching graphs
Question is to sketch the graph y=(x-3)^3 +2x which seems harmless enough. However when I find the derivative to identify any stationary points the resultant quadratic has no real solutions. I assume this means it must have no stationary points? But when I sketch in my calculator it seem to have a point of inflection. Any reason why this might be?
Derive it again, and you`ll see it has a point of inflection at x=3. (where
what`s more, the point is a rising point of inflection since:
for => concave down,
and for => concave up.
(but also, the 1st derivative with a point either side of x=3 - say x=2 and x=4, is +ve in both cases, indicating a function increasing at every point{with the exception of x=3, of course} )
All that remains is to find the (single) point at which the graph crosses the x axis.
(I`d suggest making a plot of y=(x-3)^3 and y= -2x on the same axes, and using the Newton-Raphson method to find the root.
Think of a suitable value to use as an iterant (seed)
Unparseable latex formula:
)f ' ' (x)=0
what`s more, the point is a rising point of inflection since:
for => concave down,
and for => concave up.
(but also, the 1st derivative with a point either side of x=3 - say x=2 and x=4, is +ve in both cases, indicating a function increasing at every point{with the exception of x=3, of course} )
All that remains is to find the (single) point at which the graph crosses the x axis.
(I`d suggest making a plot of y=(x-3)^3 and y= -2x on the same axes, and using the Newton-Raphson method to find the root.
Think of a suitable value to use as an iterant (seed)
(edited 10 years ago)
Original post by Europa192
Question is to sketch the graph y=(x-3)^3 +2x which seems harmless enough. However when I find the derivative to identify any stationary points the resultant quadratic has no real solutions. I assume this means it must have no stationary points? But when I sketch in my calculator it seem to have a point of inflection. Any reason why this might be?
A point of inflection can have any gradient. It is simply a point where
So when asked to sketch a a cubic, or higher degree curve it is always a good idea to check the second derivative as well as the first.
Original post by brianeverit
A point of inflection can have any gradient. It is simply a point where
So when asked to sketch a a cubic, or higher degree curve it is always a good idea to check the second derivative as well as the first.
So when asked to sketch a a cubic, or higher degree curve it is always a good idea to check the second derivative as well as the first.
Oh ok, thank you. I didn't realise that was the definition of a stationary point.
Original post by Europa192
Oh ok, thank you. I didn't realise that was the definition of a stationary point.
It isn't!
A stationary point is one where dy/dx = 0. This could be a max, min or a point of inflexion - you need info from the second derivative to determine which one.
Also note that you can have a point of inflexion that is not a stationary point e.g. y = sin x at x = 0.
Original post by davros
It isn't!
A stationary point is one where dy/dx = 0. This could be a max, min or a point of inflexion - you need info from the second derivative to determine which one.
Also note that you can have a point of inflexion that is not a stationary point e.g. y = sin x at x = 0.
A stationary point is one where dy/dx = 0. This could be a max, min or a point of inflexion - you need info from the second derivative to determine which one.
Also note that you can have a point of inflexion that is not a stationary point e.g. y = sin x at x = 0.
Oh yeah; sorry I meant point of inflection >.<
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