Post by: lshak on July 06, 2011, 11:34:36 am
My teacher sucks and I have no idea how to do these! haha they are probably really simple but range/domain has always confused me -_-
(http://i54.tinypic.com/5uqyj8.jpg)
(http://i51.tinypic.com/33eln9s.png)
(http://i54.tinypic.com/2l9bo1.png)
(http://i51.tinypic.com/10p7jgg.png)
I wouldn't have a clue how to do these on my CAS either :-\
Post by: pi on July 06, 2011, 11:38:30 am
5. Clue, the 'cusps' indicate that it is a magnitude graph. Looking a the y-int (which would be (0,-a) if the graph was not a magnitude), it must be option C.
6. Semi-circle. therefore E.
7. Inverse. You'd get D. Thanks moekamo, my bad :(
Post by: b^3 on July 06, 2011, 11:42:02 am
so the equation is |x^2-a| so asnwer is c.
Edit: Beaten, Rohitpi, so fast :)
Post by: moekamo on July 06, 2011, 11:46:58 am
6. this is a the lower half of a semicircle of radius 2 and center at (0,0). hence the range is [-2,0]. so E
7. range of f inverse = domain of f. so ran = [1,3]. so D
EDIT: Beaten, except Rohitpi you got the range of f, not f inverse! im pretty sure its D :P
7. Same as above. You'd get C.
Post by: pi on July 06, 2011, 11:51:17 am
EDIT: Beaten, except Rohitpi you got the range of f, not f inverse! im pretty sure its D :P7. Same as above. You'd get C.
+1, I needed to expand my window ;D Couldn't see the whole question
Post by: pi on July 06, 2011, 12:03:32 pm
woops
It happens. :)
Post by: lshak on July 06, 2011, 12:12:33 pm
Uhmm what would this one be? :P
For which of the following would an inverse function exist regardless of the domain?
A. parabola
B. truncus
C. circle
D. hyperbola
E. absolute function graph
Post by: moekamo on July 06, 2011, 12:24:32 pm
Post by: brightsky on July 06, 2011, 12:30:17 pm
Post by: lshak on July 06, 2011, 12:49:18 pm
Edit/ Okay what about this one?
(http://i52.tinypic.com/28ic48k.png)
For (a)...Is the domain just [-3,3] ?
For (c) is it 3 +or- squareroot k (I just guessed )
And for (d) is it k>0 ??
....And I have no clue how to do (b) :P
Post by: hairypenis on July 06, 2011, 06:24:18 pm
Post by: xZero on July 06, 2011, 06:37:11 pm
b) [-3+k,3+k]
c) +or- sqrt(9-k^2)
d) k>3
Post by: taiga on July 06, 2011, 06:55:11 pm
i think i got the same test as you! i haven't got my result back but i had no idea how to do that question ^^^either lol
That's because you're the same person? :S
I personally don't mind if you just post "bump" to get your question noticed.
PM me if I'm wrong, if not let me know which one of your (3) accounts you'd like to keep :)
You're not trying to do anything bad so obviously just take it as advice not to do it again :P
------------------------------------------------------------------------
This graph is a a circle with radius 3. centre at (0,k). hence the y values of the range are K+3 and k-3, the domain is from [-3,3]
edit: xZero has it my bad :P
Post by: kamil9876 on July 07, 2011, 12:00:21 am
Post by: xZero on July 07, 2011, 01:43:21 am
Interesting, according to the rest of the world in question 7 f would not have an inverse (a technicality, need to change the codomain) but I guess in vce it does. Fun Fact.Was trying to figure out why cant it have an inverse but failed miserably :( don't see anything wrong with R as the codomain, care to explain? thanks
Post by: kamil9876 on July 07, 2011, 10:49:56 am
Just saying that according to the mathematical community it wouldn't. Check the definition here. Wiki also uses this definition here and it even says that:
Quote
When using codomains, the inverse of a function ƒ: X → Y is required to have domain Y and codomain X.
But in question 7, R is "too big" for a codomain. If instead that question had the codomain as [2,6] then it would be fine according to this definition.
It's a better definition in higher mathematics when you're doing more complicated things. But since in vce you are only interesting in sketching graphs etc. I guess we don't need such a definition.
Post by: xZero on July 08, 2011, 03:35:37 am
Post by: kamil9876 on July 08, 2011, 11:56:08 am