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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: TheJosh on August 04, 2009, 09:39:12 pm

Title: Sketching f'(x) given f(x) and vise versa
Post by: TheJosh on August 04, 2009, 09:39:12 pm

Could someone please explain how to draw f(x) graphs from f'(x) graphs would be much appreciated

i'm so confused!!!

Title: Re: Sketching f'(x) given f(x) and vise versa
Post by: Flaming_Arrow on August 04, 2009, 09:40:42 pm

please elaborate. is the fuction given? or is it just a graph

Title: Re: Sketching f'(x) given f(x) and vise versa
Post by: TonyHem on August 04, 2009, 09:47:48 pm

Hmm nvm not that great at it myself.

Title: Re: Sketching f'(x) given f(x) and vise versa
Post by: Edmund on August 04, 2009, 09:50:44 pm

I think he means how to draw the derivative graph for a function, and the opposite.

For a f'(x) graph, whatever below the x axis corresponds to a negative gradient on the original f(x) graph.

And anything above the x axis of a f'(x) graph corresponds to a positive gradient on the f(x) graph.

Any point on the x axis corresponds to a zero gradient.

Title: Re: Sketching f'(x) given f(x) and vise versa
Post by: TrueTears on August 04, 2009, 09:53:38 pm

You can remember some general rules

When you differentiate a quartic you get a cubic graph.

A cubic - > parabola

and so on.

Title: Re: Sketching f'(x) given f(x) and vise versa
Post by: /0 on August 04, 2009, 09:54:24 pm

Just going through the motions:
Where f'(x) = 0 , f(x) has a stationary point.
Where f'(x) < 0 , f(x) has a negative gradient.
Where f'(x) > 0, f(x) has a positive gradient.
Where f'(x) has a local min/max, f(x) will have a point of inflexion.

Title: Re: Sketching f'(x) given f(x) and vise versa
Post by: TheJosh on August 05, 2009, 03:51:54 pm

 ohhhh ok i think i see it now

thanks guys appreciate the help ;)