2000 HSC 9 b) help (1 Viewer)

[Shifty]

Can someone please explain this graph question as easy as they can because everyone who has explained it has done a pretty average job.
The question is from 2000 HSC. Question 9 b)


(Attachment: Graph)


The above diagram shows a sketch of the gradient function of the curve y = f(x).
Draw a sketch of the function y = f(x) given that f(0) = 0

 

[Seraph]

all i can say is

when going back from f'(x) to F(x)

now unless ive got this muddled up

where you got ya negative gradients , this means that the curve is Below the axis on y and where it is positiev gradient it is above , .. also im pretty sure min/max are POI on f(x) and also since f(0) this means that we know that it starts from the origin

pretty lousy definition but i need some clarifying as well

 

[acmilan]

To find turning points you look at where f'(x) = 0 (in this case at x=1 and 3), then you look at the sign of f'(x) for these TP values both just before it and after it. This will give their nature.

For example at x = 3 the graph is + before and - after the point x=3 and hence is a max tp for y = f(x)

 

[Shifty]

But how far up do we know where to sketch it....etc??:S

So is the reason for at x=1 being a horizontal point of inflexion because its is 0 before and 0 after?

And why does 2 become a point of inflexion

Please bare with my stupidness....thanks

The solution:

 

[jumb]

From the information on the graph, I discovered that the graph is:

y = -2/3(x-1)^2 * (x-3)

The integral of which is

y = 2*x - (7*x^2)/3 + (10*x^3)/9 - x^4/6

(c = 0)

Both of these graphs are shown on the attachment.

 

[withoutaface]

Well you start with f(0)=0 and put that point in.

There is a point of inflection at x=1, because there is a turning point in f'(x) there. This point of inflection is also stationary because f'(1)=0

You then have another point of inflection at x=2, and then a maximum at x=4. So long as your graph has these basic features nothing else matters.