At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
Make a table of values for each function. To do that, come up with some values for x and calculate f(x) and g(x) f(x) = |x| x | f(x) ---------- -2 | 2 -1 | 1 0 | 0 1 | 1 2 | 2 Use the above points to graph f(x). Then do the same for G(X)
@samnatha r u there?????/
yes sorry was just tying to work it out
i then have to solve |x| = | 2x-3|
is anyone there ?
No. You don't need to solve that equation. Just make two tables of points and graph both on the same x-y graph
oh ok thanks theres another one which is |2x-3| < |x|
Sorry, I misread your post. When you said you have to solve |x| = |2x - 3|. I thought you meant that you needed to solve that to get the graph. After you graph both f(x) and g(x on the same graph, see where the graphs intersect. That's the solution to|x| = |2x - 3|
ok thanks i got that bit but i'm not too sure on the second bit
The soulition to |x| = |2x - 3| is where the two graphs intersect. There are two points.
(1,1) and (3.5, 3.5) thats what i got on my graph is that right ?
The solution to |2x - 3| < |x| is the x values for which the value of g(x) is less than the value of f(x). If you look in graph, see where the y vaslue of g(x) is lower than the y value of f(x).
I think it's (1,1) and (3,3)
ok my graph is probs a bit diff but thanks for your help :)
god the pressures of leaving cert math