Re: 回复: Re: 回复: Re: Maths Extension 2 Trial Exam thoughts- 2008
Arowana21 said:
Lets say you have a graph
f(x)= 1- e^x
and the question asked you to draw f(x^2)
what would be the solution
This is a graph of composite functions.
The best way to approach these is to do a bit of inspection:
Features of y = 1 - e
x: (You could do the same thing if the graph is given rather than the equation)
- At x = 0, y = 0
- As x approaches ∞, y approaches - ∞
- As x approaches - ∞, y approaches 1
Features of y = 1 - e
x²: (just replace x with x² and note additional restrictions)
- At x² = 0, y = 0 (i.e. at x = 0, y = 0)
- As x² approaches ∞, y approaches - ∞ (i.e. As x approaches ∞, y approaches - ∞ and as As x approaches - ∞, y approaches - ∞)
- As x² approaches - ∞, y approaches 1 => ignore this as x² is always non-negative (an additional restriction)
*Another thing to note is that it is an even function, so there is y-axis symmetry
*There is a turning point at the origin in this case as the derivative equals zero at that point
[In general x-coordinates of turning points are conserved for composite functions: i.e. differentiate f(g(x)) w.r.t x gives f'(g(x)).g'(x) so the x-coordinates of the turning points of g(x) are conserved since g'(x) = 0 is a solution to f'(g(x)).g'(x) = 0]
Using the above information, you can deduce the graph...(it should look like a negative "parabola" from the origin)