3u hsc 1994 (inverse trig) (1 Viewer)

[+:: $i[Q]u3 ::+]

okay u know how when u find the domain of an inverse function it's got to be the largest one-one domain???

how come with the inverse function question in the 3u hsc paper 1994... (it's like some cubic graph...) the answer that they give for the domain is the middle bit of the curve? doesn't the bit to the right or the bit to the left have a larger domain for which the inverse can exist?

have i totally lost all u ppl? ah well.. find the 3u paper and u'll see what i mean.

 

[ezzy85]

the question says that the origin has to be included. the other domains dont include the origin.

 

[+:: $i[Q]u3 ::+]

oh right thanks.. lol.. dammit i sound like an absolute lostie ^^