Sketching graphs Watch

Europa192
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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?
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Hasufel
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Derive it again, and you`ll see it has a point of inflection at x=3. (where f ' ' (x)=0)

what`s more, the point is a rising point of inflection since:

for  x<3, f '' (x<3) is -ve => concave down,

and for x>3, f '' (>3) is +ve => 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)
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brianeverit
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(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 \frac{d^2y}{dx^2}=0
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.
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Europa192
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(Original post by brianeverit)
A point of inflection can have any gradient. It is simply a point where \frac{d^2y}{dx^2}=0
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.
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davros
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(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.

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Europa192
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(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.

Oh yeah; sorry I meant point of inflection >.<
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