Can someone please help me with this!
There are approximately five times as many magpies as currawongs in a certain area.
If the population of currawongs increases at a rate of 12% per annum while that of
the magpies decreases at 6% per annum, find how many years must elapse before the
proportions are reversed, assuming the same rates continue to apply.
(Q9 EX 5I)
Thanks!
Welcome to the forums!
Let the starting number of magpies be \(5x\), and the starting number of currawongs be \(x\). Then, after \(n\) years, the number of magpies will be \(5x(1-0.06)^n\) and the number of currawongs will be \(x(1+0.12)^n\). We want to find when the number of currawongs will be 5 times that of the magpies or more - ie what is the smallest value of \(n\) that satisfies the inequality \(\frac{x(1+0.12)^n}{5x(1-0.06)^n} > 5\) ie. solve this inequality for \(n\) - \(\left(\frac{1.12}{0.94}\right)^n > 25\). Try this out
I feel like Ive been spending all afternoon/evening trying to understand something simple.
You haven't, I probably started with too many assumptions
I don't see how you got (a+b)/2. Or why that's true
The largest domain for which the parabola has an inverse lies between positive/negative infinity and the vertex. The vertex is the maximum if the parabola is a 'cap' or a minimum if the parabola is a 'cup'. More importantly for our purposes, the parabola is symmetrical across the vertical line that passes through the vertex. Since this is the case, its x-coordinate must be halfway between each of the roots a and b, hence (a+b)/2. The horizontal line test (if a function has some real y value, any real y value for which a horizontal line through that y value cuts through the function twice, it does not have an inverse across its entire domain - you have to restrict the domain for it to have an inverse) fails because of this symmetry. This symmetry automatically gives you the second point that fails the horizontal line test - in fact the parabola fails this test everywhere in the range of the function (except for the vertex).
Also, I can see how a square root graph has a 'root'. If a parabola has a root, would it not be one root with two 'sprouts', not two (a and b). Im not sure why you pointed this out specifically
I didn't use root in inverted commas because roots weren't intentionally metaphorical like 'branches' (which is a commonly used term, I think? but not formal). Roots are just another way of saying the solutions of the parabola, or if it suits you better the x-intercepts of the parabola. Hopefully things make more sense with this clarification - all parabolas have two roots (either two distinct roots, two duplicate roots or two complex roots (but this is a whole different story)).
No idea what that tells me about b.
You can put \(x^2+4x\) into the form \((x-a)(x-b)\) by factorising it. This will tell you that the roots of the equation are 0 and 4.
Since b is the smallest value for which \(g\) has an inverse, it follows that b has the same x value as the vertex ie. (a+b)/2 = (0+4)/2 = 2.
Also, all I know how to do is to substitute y for x and do the algebra bit. But I didnt know how to isolate y and so got this far:
f^-1(x)=y(y + 4)
The easiest method is probably completing the square.
Depending on your domain restriction (assuming it is restricted such that the domain is the largest for which g has an inverse) you now have to choose a sign for the root because choosing both means that this is not a function. For 'cup' parabolae if you choose the one that has an open interval in the positive x direction you choose the positive root, otherwise you choose the negative root. For 'cap' parabolae the opposite is true.
Hope this makes more sense