Recent content by Tevion

so do we make it? for the inverse the image set (range) (0,1/2pi] for the domain (1.5, infinity)

your functions should look like this:

2x^2

the range is the function is 0 < x < 1/2pi Thanks for helping

i hear what you're saying bud, but any advice how to take that further? I don't have a clue :(

ok Thanks - so how do i go about do that? basically from my attachment, i need to explain why fr has fr^-1, stating the domain and image set a hand would be great with it please

how does this explain that is has an inverse function

here we are plotted: so how do we get the image set and domain

Ok, i need to explain why this function has an inverse function. fr^-1 fr(x) = 1.5/sinx and state its image set & domain. Is that (0,1/2Pi] Thanks :)

8 x (2 x^2+1) (6 x^2+1)

2 (2 x^2-1)

now its a bit trickier and would like some help please. rather than the angle being 65 degrees, but is x in radians where 0 < x < 1/2pi. then Nr depend on the value of x. the Nr from the lift have mag. fr(x) N. Assuming domain of F fr is (0,1/2pi] 1) we need to use the triangle of...

Well i won't turn down assistance, and advice, so please go on :)

got it :D

Did that help?