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[SxC]

This is like a really simple question i think but i dont know how to get it
its Q 5(d) from 1996... if uz cant draw the picture could uz just explain like how it workz

 

[JayWalker]

This is like a really simple question i think but i dont know how to get it
its Q 5(d) from 1996... if uz cant draw the picture could uz just explain like how it workz

Could you please post the question???

 

[SxC]

For all x in the domain 0{ x } 4, a funtion g(x) satisfies g'(x){0 and g"(x){0

{ = greater or equal
}=smaller or equal

 

[JayWalker]

Im ASSUMING your talking about the g(x) question..

Well g(x) is a function and they want you to sketch a possible graph..

They give you this information

g^'(x) < 0
g^''(x) < 0

well g^'(x) is the gradient function, so the curve has to be negative from 0 to 4...

And g^''(x) is concavity, so its gotta be concave down... So it would sorta look like you started on the Y axis, and chucked a ball and traced it path.. Do you understand why??

Like if you drew a circle, and only took the top right hand corner...


Understand?

Rep me

 

[SxC]

like when it says g"(x)<0 does that mean that it concaves down
and g"(x)>0 means up ???

 

[JayWalker]

like when it says g"(x)<0 does that mean that it concaves down
and g"(x)>0 means up ???

Yea yea you got it!

Imagine your flying an aircraft...
You push forward (convention is forward is down...) and you start decending
Pull back and you start ascending.. its like the 'curve' of the circle

g^''(x) is an expression for the rate of change of your gradient...

kewl?

 

[SxC]

Yeah thanx bro but how bout g'(x) < 0 how do u know that it starts above the gaphh

so g'(x)<0 starts above and g'(x) > 0 starts below

 

[JayWalker]

Yeah thanx bro but how bout g'(x) < 0 how do u know that it starts above the gaphh

Assuming that you mean "above the gaphh" is above the x axis, i dont.. but its a possible representation of that function being they did not give an "Anchor Point"

so g'(x)<0 starts above and g'(x) > 0 starts below

No, i just gave you one POSSIBLE graph... g^'(x) is the gradient,,
i.e. g^'(x)>0 then its line going up to the right,, g^'(x)<0 then its line going up to the left...

this question, unless it said that g(x) goes thru the point (x,y), you can draw it where ever you like (UNLESS it gives you the domain). For this question ,it gives you a domain, but you cna start either at +ve 9999999 or -ve 99999 it doesnt matter... Its only the gradient.. They didnt say where g(x) was and hence the inclusion of the word "possible"

 

[SxC]

okay thx bro for ur help i could do it now

 

[JayWalker]

No problem

rep me

 

[SxC]

hey bro rep me too even though i didnt help u particularly

 

[JayWalker]

done, do they just add on automatically?

 

[SxC]

yeah i guess ok now how do u close this