Graphing derivative functions - help? (1 Viewer)

[Fiction]

Hi~
I'm getting stuck on drawing graphs of derivative functions, and was wondering if someone could be sweet enough to tell me what the graph of a derivative function would look like when its primitive has a turning point and stationary point. Also the attached image - did I draw it right?
Thanks C:

 

[astroman]

https://www.youtube.com/watch?v=Gbtma_UQpro

 

[Flop21]

What is a primitive?

 

[Fiction]

Hi~
I'm getting stuck on drawing graphs of derivative functions, and was wondering if someone could be sweet enough to tell me what the graph of a derivative function would look like when its primitive has a turning point and stationary point. Also the attached image - did I draw it right?
Thanks C:View attachment 31401

https://www.youtube.com/watch?v=Gbtma_UQpro

Thanks C: I would've preferred specifics tho lol

What about the second half of moi question? ;o

What is a primitive?

LOL
It's what you get when you intergrate.
dw about ^, I doubt you'll get an assessment on it soon if you don't know what it is

 

[Fiction]

https://www.youtube.com/watch?v=Gbtma_UQpro

urgh vid made me more confused. at 10:53, she has the complete graph - why is the derivative of the middle of her horizontal line between her global min and max, touching the x axis? :s

EDIT:
NVM, realised it was a horizontal point of inflexion. My bad.

 

[astroman]

urgh vid made me more confused. at 10:53, she has the complete graph - why is the derivative of the middle of her horizontal line between her global min and max, touching the x axis? :s

EDIT:
NVM, realised it was a horizontal point of inflexion. My bad.

video is da best, u should be able to understand it all from this.

 

[photastic]

Hi~
I'm getting stuck on drawing graphs of derivative functions, and was wondering if someone could be sweet enough to tell me what the graph of a derivative function would look like when its primitive has a turning point and stationary point. Also the attached image - did I draw it right?
Thanks C:View attachment 31401

Perfect. Quick notes:
To find derivative functions:
Increasing f(x) = Above the y axis for f'(x)
Decreasing f(x) = Below the y axis for f'(x)
Stat pts f(x) = Roots for f'(x)
POI f(x) = Stat pt for f'(x)
HPOI f(x) = Stat pt + Root for f'(x)
Consider other conditions such as f(0) = 0 and ignore any roots of the original graph.

When finding the original graph (primitive) from its derivative, simply work backwards.

 

[Fiction]

Perfect. Quick notes:
To find derivative functions:
Increasing f(x) = Above the y axis for f'(x)
Decreasing f(x) = Below the y axis for f'(x)
Stat pts f(x) = Roots for f'(x)
POI f(x) = Stat pt for f'(x)
HPOI f(x) = Stat pt + Root for f'(x)
Consider other conditions such as f(0) = 0 and ignore any roots of the original graph.

When finding the original graph (primitive) from its derivative, simply work backwards.

"You must spread some Reputation around before giving it to aarondapho again." :c

Thanks btw c: What do you mean by consider other conditions such as f(0) = 0 and ignore any roots of the original graph?

 

[photastic]

"You must spread some Reputation around before giving it to aarondapho again." :c

Thanks btw c: What do you mean by consider other conditions such as f(0) = 0 and ignore any roots of the original graph?

Some questions might give you extra info like f(0)=0 means when x = 0 f(x) = 0 essentially meaning it passes the origin. Other conditions include what coordinates the function passes through at.

Also, when you want to find the derivative function, you will realise that the x intercepts (roots) of the original function do not necessarily influence the derivative graph.

 

[funnytomato]

Hi~
I'm getting stuck on drawing graphs of derivative functions, and was wondering if someone could be sweet enough to tell me what the graph of a derivative function would look like when its primitive has a turning point and stationary point. Also the attached image - did I draw it right?
Thanks C:View attachment 31401

yes, just note that the point of inflexion of f(x) ----> MAX or MIN (depending on the sign of gradient of f(x)) of f'(x)

In your question, it is MIN because the slope is most NEGATIVE at that point.

Remember that derivative function f'(x) simply represents the gradient of f(x), and gradient is just rise/run (see post below )

so you can always just draw the tangent(or what appears to be a tangent) on f(x), and see how the gradient function f'(x) changes

 

[funnytomato]

here we have y=x^2 and its derivative y'=2x

(generally the equation of the function will NOT be given, but let's look at this simple example to understand what's actually happening)

you can choose any point on f(x), draw the tangent , and find out its rise and run (here I've used numbers just so that it's more apparent, but we just need whether it's getting larger/smaller or approaching 0 )

 

[funnytomato]

"You must spread some Reputation around before giving it to aarondapho again." :c

Thanks btw c: What do you mean by consider other conditions such as f(0) = 0 and ignore any roots of the original graph?

I think what he means is that for a given f(x), graph the primitive function F(x). Then you always have a family of curves(F(x)) having the same gradient.

For example, if given a graph like y=2x, then its primitive could be y=x^2(like the above diagram, backwards),
but y=x^2+1 , y=x^2- pi ... all fit our criteria of having derivative of 2x.

Hence to graph F(x), the question needs to specify a boundary condition
like it passes through the origin (i.e. F(0)=0), which means that F(x)=x^2
Had you been given a different boundary condition, say F(0)=1, then you'd have a different primitive ---> F(x)=x^2+1

Again remember the question will not give you the equation explicitly, but what you're doing is essentially the same thing as above