help with 2000 2u paper please =) (1 Viewer)

[danif]

heya,
can someone help me with question 9b) from the 2000 paper?
what is the general way of doing these kinds of questions? and how do you do them if they give you the original function and ask you to graph the gradient function using the graph of the original function (if you get me...) ?

thanks heaps =)
dani

 

[minicam]

When the gradient function touches the x-axis, its a turning point for the normal function....if its negative gradient at that point then its a maximum turning point and vice versa.
In the gradient function when it touches the x-axis and is also a turning point, then in the normal function it will be a horizontal point of inflection.
When theres a turning point in the gradient function its just a point of inflection.
So for the question 9b in the 2000 paper, it begins at 0, then at x=1 its a horizontal point of inflexion above the x-axis, at x=2 is a point of inflexion and x= is a turning point where it then begins to decrease.
Hope that helps!!

 

[Xayma]

Where it hits 0 is a stat point.

Where there is a stat point is a point of inflexion.

It pases through the origin

 

[smallcattle]

i still dont understand how to draw f(x) from f'(x)...

can someone elaborate abit more?? thx...

 

[Seraph]

evidently from f(x) to f'(x)

turning points are intercepts on f'(x)
furthermore POI on f(x) will be turning points on f'(x)

this being said plot these points know this ... if the curve of f(x) is a negative gradient it is below the axis on f'(x) , hence when it is a positive gradient on f(x) it is above the x axis on f'(x)

 

[Tommy_Lamp]

It's easier to do this when you know what each derivative stands for, i.e. f"(x) > 0 = Concave Up etc.