Derivative graph and the graph of the original function

[kLPantera]

Homework Statement

When given the graph of f '(x) the graph sort of oscillates above the x-axis and the graph of
g '(x) which starts in Quadrant III sort of curves up, goes through point (0,0) keeps going up then begins to even out.

Homework Equations

How can you tell how many solutions there can be when f(x) = 0? How can you tell how many solutions there can be when g(x) = 0? and If g(x) has 2 solutions where would they be?

I don't seem to understand this.

Any help would be appreciated

 

[Coto]

You need to have some knowledge of the form of the equations you're dealing with. Do you have a functional form for the derivatives of f'(x) and g'(x)?

If you don't know the forms of these equations, then I'm not sure if what you're asking is possible.

 

[FLUndergrad]

well if f'(x) is never above the x-axis then the slope is never negative, so there could only be one solution for f(x)=0 right? For g(x)=0 if there are two solutions then one solution would have to be for when x<0 and x>0 since the slope only changes it's sign after the graph crosses the origin. but that's only if g(x) crosses the x-axis before x becomes > 0, or else it can't go back across the x-axis once the slope changes. That's just how I visualized it, hope that helps a little.

 

[kLPantera]

There are only 2 graphs: f'(x) and g'(x).

I understand the one for g(x) now however...

For the f'(x) graph, f'(x) is always above the x-axis. That means the slope is always positive?

 

[brydustin]

If the functioon f ' (x) is strictly positive (above the x-axis) then this means that the slope of f (x) is always positive. So the function f (x) can have at most one (real) root, or none as is the case with f (x) = e^x.

There are only 2 graphs: f'(x) and g'(x).

I understand the one for g(x) now however...

For the f'(x) graph, f'(x) is always above the x-axis. That means the slope is always positive?