Hi, can someone explain me how to do those question giving you a y=f(x) graph and asking you to sketch the f'(x) of it and vice versa. Is there a set of rules to follow in order to draw these? because I keep getting them wrong every time.
-
Looking for HSC notes and resources? Check out our Notes & Resources page
Graphing f(x) and f'(x) (1 Viewer)
- Thread starter Newbit
- Start date
GoldyOrNugget
Señor Member
- Joined
- Jul 14, 2012
- Messages
- 583
- Gender
- Male
- HSC
- 2012
Stationary points of f(x) become x-intercepts of f'(x).
Regions of positive gradient on f(x) are above the x-axis of f'(x).
Regions of negative gradient on f(x) are below the x-axis of f'(x).
Horizontal points of inflexion on f(x) become stationary points of f'(x).
Vertical points of inflexion on f(x) become vertical asymptotes of f'(x)
Regions of positive gradient on f(x) are above the x-axis of f'(x).
Regions of negative gradient on f(x) are below the x-axis of f'(x).
Horizontal points of inflexion on f(x) become stationary points of f'(x).
Vertical points of inflexion on f(x) become vertical asymptotes of f'(x)
Whats the difference between a horizontal and vertical POI (hopefully there is a better explanation than just "one is horizontal the other is vertical.."
Jashua_Long
Banned
- Joined
- Oct 15, 2012
- Messages
- 58
- Gender
- Undisclosed
- HSC
- N/A
Horizontal ----> both f' and f'' are zero. The graph flattens out and inflects.
Vertical ---> f'' =0, but f' does not equal zero.
They have different shapes and should be able to draw them
When you think about them graphically, all of the points mentioned by GoldyorNugget should make quite clear sense.
For example, check this link:
http://www.teacherschoice.com.au/Maths_Library/Calculus/stationary_points.htm
The second graph on that page shows a horizontal point of inflection.
The third graph on that page shows a normal point of inflection (or vertical point of inflection) at approximately x=+0.6
Last edited:
GoldyOrNugget
Señor Member
- Joined
- Jul 14, 2012
- Messages
- 583
- Gender
- Male
- HSC
- 2012
Eek I messed up a bit: all non-vertical POIs become stationary points. Horizontal POIs become stationary points that bounce on the x-axis.
A POI is a change of concavity. When you change from concave-down to concave-up, the gradient has been getting steeper downwards for previous x values, but now it's going to become less steep as we transition to concave-up. 'Gradient becoming steeper' manifests itself on f'(x) as a region of increasing y-value, and 'gradient becoming less steep' is a region of decreasing y-value.
A horizontal POI is when the gradient is positive, then flat, the positive again (or negative, flat, then negative). If a skier was at
the top (on either side), he'd be going downhill, then he'd be on flat ground for a bit, then downhill again. These become stationary points on f'(x) at y=0 because the gradient transitions from positive->zero->positive or negative->zero->negative.
A vertical POI is when the gradient is positive, then vertical, the positive (or negative, then vertical, then negative). Because the gradient is effectively infinite at these points, they become undefined on f'(x), so there are asymptotes at these x-values.
A POI is a change of concavity. When you change from concave-down to concave-up, the gradient has been getting steeper downwards for previous x values, but now it's going to become less steep as we transition to concave-up. 'Gradient becoming steeper' manifests itself on f'(x) as a region of increasing y-value, and 'gradient becoming less steep' is a region of decreasing y-value.
A horizontal POI is when the gradient is positive, then flat, the positive again (or negative, flat, then negative). If a skier was at
the top (on either side), he'd be going downhill, then he'd be on flat ground for a bit, then downhill again. These become stationary points on f'(x) at y=0 because the gradient transitions from positive->zero->positive or negative->zero->negative.
A vertical POI is when the gradient is positive, then vertical, the positive (or negative, then vertical, then negative). Because the gradient is effectively infinite at these points, they become undefined on f'(x), so there are asymptotes at these x-values.