given y = f'(x), sketch y = f(x) (1 Viewer)

[ToO LaZy ^*]

help please?..i can do the ones without the asymptote..but..not these.
i have the solution, but it doesn't explain anything, it just shows you the graph
i will post the solution up later.
thx

 

[ngai]

as x -> 3, y' -> infinity
so as x -> 3, gradient of y -> infinity
ie. graph of y = f(x) is vertical at x = 3
the rest u can do?

 

[Xayma]

There is also stat points at x=2 and x=-2, with points of inflexion at x=-2 and x=-1.

 

[BlackJack]

Okay, on the left side of x=3, the graph y=f(x) does goes upwards to infinity, since the gradient is very large and positive.
Look at the right side of x=3. The gradient is still a very large positive number. This means the graph y=f(x) is coming up from negative infinity.

Compare this part of the graph with the asymptote of xy=-1; the two are pretty much the same.

 

[ToO LaZy ^*]

hmm..

ngai:..yep, got that bit

xayma: so whenever y = f'(x) has an x-intercept, those points will be stationary points?
and whenever there are stationary points, there are inflexions?
ok, that seems right

blackjack: ok, makes sense

what if you were to write down the equation?...that would be hectic..!

EDIT: anyone wanna do a rough sketch..?

 

[Xayma]

Yes, as f(x) has a stat point where f'(x)=0.
Where a stat point isnt a horizontal point of inflexion (I pray) that it should be a point of inflexion in y=f(x)

Nature of stat points:
At x=-2, slope is always negative, therefore it is a point of inflexion however coming down from infinity, then it goes through another point of inflexion and then a stat point (minimum) and back up to infin.

Dont worry about where the x axis lies, because the constant will change that.

So you could keep it all in the 2nd and 4th quadrants etc.

 

[ToO LaZy ^*]

for real..
i'll post up the solution now..