Help Please - Urgent, Assesment due tomorrow (1 Viewer)

[crippled_kungfu]

Hi guys im in a spot of trouble here, i've got a maths sheet due tomorrow and its worth 10% of our school based assesment. However theres one annoying question i cant do so any help would be greatly appreciated. Ive added the question as a small (600kb or so i think) gif file so you can view it there any help is greatly appreciated.

Basically i just need to know how to find the equation of f(x) then i should be able to do the rest but if u can help with all the answers that would be awesome thanks.

Anyways thanks again any help is definately welcome and please respond ASAP.

 

[crippled_kungfu]

damn it wont let me add attachments hmm the question is a differentiation/integration one where ur given a graph and its of the derivative of a function f'(x) and then u gotta find the turning points, inflexion point etc of the graph of f(x) it doesnt give u the equation of either u gotta work them out. If you think you may be able to help me email me at stephen_bradley@iinet.com.au and i'll email you the question thanks alot, in the meantime i'll keep tryin 2 get this attachment to work on the forum

 

[FinalFantasy]

i'll help ya, but y can't u type out da q. here?

 

[acmilan]

well you know that turning points occur when f'(x) = 0 so you find these points and then see what the sign is on either side of these points to get the nature of the turning point.

 

[crippled_kungfu]

Question

Hey guys decided as i couldnt get attachment to work id upload it onto a geocities site, to view the question simply go here:

http://www.geocities.com/bodgie316/question11.gif

 

[FinalFantasy]

so which one can't u do?

 

[crippled_kungfu]

i cant do any of them

 

[FinalFantasy]

o i thought u did dem cuz u had ticks on dem lol

 

[Lhyviathan]

(i) x = - 2, relative minimum
(ii) x = 3
(iii) -2 < x < 3, x > 3
(iv) well... it's not that hard, and I can't do it here.

EDIT: not sure about the definition of "increasing"... does that mean it is positive and accelerating (f"(x) > 0), or simply where the gradient function is positive (f'(x) > 0)?

 

[FinalFantasy]

for first one just use the idea that when u get a turning point on f(x) when u put f'(x)=0
so x=3 would be ur f'(x)=0

 

[FinalFantasy]

hmm maybe im wrong! haha, im bad at graphs

 

[Lhyviathan]

At x = 3, you have a horizontal point of inflexion, which I don't think is classified as a "turning point".

At x = - 2, however, you have a minimum turning point.

 

[FinalFantasy]

ooer ic
--------- 10 letters

 

[crippled_kungfu]

how did you find that theres a point of inflexion at x=3 and a min turning point at x=-2

 

[Lhyviathan]

I sketched the graph of y = f(x), given y = f'(x).

Basically, just drawing the original graph, given the graph of the derivative.

Whenever you see on the graph of the gradient function (that's y = f'(x)):

- BELOW zero... that's where the gradient of the original graph (y = f(x)) is negative.
- ZERO... that's where there gradient of the original graph is ZERO... that's where you have a STATIONERY POINT
- ABOVE zero... that's where the gradient of the original graph is positive...

Now, you just sketch it according to the critical points, and HOW positive or negative the gradient is at a given point.