Sketch f(x) from f'(x) (1 Viewer)

[_ShiFTy_]

From what i've gathered from doing one question, and looking at the answers...are these the main steps:

1) If the y = 0 , then stationarly point for f(x) curve
2) If stationairy point on f'(x) curve, then point of inflexion on f(x) curve?
3) If stationairy point AND y = 0 , then HORIZONTAL point of inflexion?

Is there anything else?

 

[joesmith1975]

Yere thats about it i think, havent done them since last year

 

[Riviet]

From what i've gathered from doing one question, and looking at the answers...are these the main steps:

1) If the y = 0 , then stationarly point for f(x) curve
2) If stationairy point on f'(x) curve, then point of inflexion on f(x) curve?
3) If stationairy point AND y = 0 , then HORIZONTAL point of inflexion?

Is there anything else?

Those seem right, you could also consider when the gradient goes from positive to negative, ie from above x axis, cutting the x axis, then going below the x axis, how fast it's increasing and decreasing.

 

[_ShiFTy_]

Just thought of a few things. What if there was a vertical asymptote on the f'(x) curve? Would this just mean that the gradient will keep on increasing/decreasing so it would look parabolic or something?
And also, if there was a vertical tangent on an f(x) curve, what would it look like on the f'(x) curve?

 

[Riviet]

Just thought of a few things. What if there was a vertical asymptote on the f'(x) curve? Would this just mean that the gradient will keep on increasing/decreasing so it would look parabolic or something?
And also, if there was a vertical tangent on an f(x) curve, what would it look like on the f'(x) curve?

If there was a vertical asymptote on the f ' (x) curve, then the original f(x) will also have the same vertical asymptote, since the gradient is increasing to infinity.
I'm not sure about the vertical tangent on an f(x) curve. I don't think this one is important since if there was a vertical tangent, then the curve would be a relation.

 

[Sober]

Just thought of a few things. What if there was a vertical asymptote on the f'(x) curve? Would this just mean that the gradient will keep on increasing/decreasing so it would look parabolic or something?
And also, if there was a vertical tangent on an f(x) curve, what would it look like on the f'(x) curve?

I believe Riviet is incorrect. These two problems are actually the same thing and one implies the other.

Consider the function of a half circle, at the endpoints the gradient becomes vertical and the derivative is asymptopical at that point.

 

[Riviet]

Sorry, wasn't thinking properly with the vertical tangents, of course the function could have a vertical tangent at its end point.

 

[Sober]

Sorry, wasn't thinking properly with the vertical tangents, of course the function could have a vertical tangent at its end point.

Not just the end point, consider f(x)=x^1/3