Hi all, I have an Engineering interview at Oxford this Monday and I am currently trying to practice for it. I see they often give a funtion and ask you to graph for it. Im just looking for general advice as to how to go about this.
Things I know to do...
- find where it cuts x and y axis
- look for turning points
In what order should I first attempt these or does it not matter?
Here is a more specific example
Take y= (sinx) / x (no idea how to write in proper maths form)
I don't know how to go about arriving at this graph...
Can someone please talk me through it if you can? I don't understand the oscillations decreasing as x increases/decreases towards infinity, and
What are limits and how can I go about calculating them? I havent covered this yet.
Thanks all, I hope someone can help with at least something
Maths Graph Drawing Help - Please Help :D Watch
- Thread Starter
- 05-12-2010 20:10
- 05-12-2010 20:27
Key things to look for are:
- where it intercept the axes (if possible)
- turning points
As you've said. Additionally:
- behaviour as x (and y) tends to positive and negative infinity
- other special features
In the sinx/x example, do you know the small angle approximation? It states that for small x, sinx is roughly equal to x, so when x = 0 sinx/x is roughly 1.
The points where it intercepts the x-axis can be found by noting what values for x make sinx = 0
Turning points can be found by deriving the function, but can't be found very easily numerically. Try plugging in a few values.
Special features: Note that sinx alternates between 1 and -1. so if we consider the graphs of 1/x and -1/x we have an envelope (we are always multiplying 1/x by a number so 1/x and -1/x 'envelope' the function, see Envelopes. Notice that the peaks appear to decrease in a similar way to 1/x and -1/x).
As x tends to positive and negative infinity, what do you notice?Last edited by Goldfishy; 05-12-2010 at 20:29.
- 05-12-2010 20:30
You could see it as multiplying sin(x) by 1/x. sin(x) just oscillates between -1 and 1, and 1/x tends to zero as x goes to infinity. In terms of limits, the limit of 1/x as x tends to infinity can be written as:
lim x->infinity (1/x) = 0.
This is the same for negative values of x, o the oscillations decrease as you move away from x=0. You could also see it as when |x| increases, you're dividing by an ever increasing value, so the oscillations of f(x) become smaller and smaller.
- Thread Starter
- 05-12-2010 20:52
Wow thanks guys, that's really a lot of help. + rep.
- 05-12-2010 20:58
...for small x, sinx is roughly equal to x, so when x = 0 sinx/x is roughly 1.
- 05-12-2010 21:26