I am having difficulties on loads of topics.
The main difficulties are on :
-Applications of differentiation
I get all of this except how to sketch a graph? You may think i am dumb and i probably am but i get conflicting messages when reading through this book.
To sketch a graph you would solve f(x) ? To find values where it crosses x and y axis. Why in the book ( Edexcel) Do they use f '(x) to find out where it crosses the axis for increasing and decreasing functions.
I understand that to find the increasing/decreasing values you use f '(x) > or < 0
But what i dont understand is the sketching of the graphs part . How comes they use the values that have been found for increasing/decreasing values.
My other problem ( BIGEST) is the trignometric functions topic.
I know the graphs , I know th C.A.S.T rule and the basis of the topic. But when it comes to solving the equations and finding values i often get stuck/wrong. Its really doing my head in so any advice on this . I would also be most greatful.
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- Thread Starter
- 16-06-2005 16:21
- 16-06-2005 16:38
When solving a trig equation in a certain range, you....
First - Extend the range.
So, if it is sin(3x) = 0.5 0<=x<=360
Find all the values of sin that = 0.5 in the range 0<=x<=1080
Then divide all those answers by 3 and that's your values of x.
I never use that cast thing. Just remember the shapes of the graphs between 0 and 360°.
To find out where the graph crosses a co-ordinate axis, let either x or y = 0.
It may well be the case that they give you the gradient of the point where it crosses the axis. In which case, do dy/dx to get the value of x when the graph has that gradient then the co-ordinates are the x value you got and y would = 0 because you are crossing the x-axis.
Also, if you only have the co-ordinates of the points where it crosses the axes, how do you know which direction the graph is going it once it has gone through that point? You'd have to differentiate to see if it goes up or down.
Finally, a good sketch shows all the stationary points on the graph.