Sketch Graph: Domain 0<x<12 | f' Increasing, f''<0

[riri]

Hi!

I'm struggling with this question. I'm supposed to draw a graph that follows and has these features:
- Domain of f is0<x<12
- f′ is increasing on (−∞,3)
- f′′ <0 on (3,6)
- f′ is Concave Up on (6, 9)
- f has infelction point at x=9!

I don't know if I can get help because this is a drawing graph question... but rightnow, I plotted the important points. I'm confused on how to draw the "f' is increasing on (-inf,3) and f"<0 on (3,6).
Is there a strategy for these types of questions? Please any help would be appreciated thankyou! :)

 

[mathmari]

Hey riri!

Since $f'$ is increasing on $(−\infty ,3)$, we have that $f$ is concave up on that interval.

Since $f''<0$ on $(3,6)$, we have that $f'$ is decreasing on that interval, that means that $f$ is concave down on that interval.

 

[riri]

Hi! Thank you! I'm still a bit confused

So I'm just stuck on how the graph is going to look. From the info, so basically, to the left of the point (3,6), would I draw a upwards graph curve to make a concave up? And to the right of point (3,6) would I draw the graph facing / going downards direction because it's concave down?

But wouldn't that make an x^3 shape which doesn't? have a concave up or down?
Sorry I'm just confused :(