I did get the domain and range (same as yours) but I didn't know the graph would be upside down to a normal inverse cos curve.
This is how i pictured it:
y = cos-1(1-x) <-- how do you sketch this? I thought i did it right, but when i checked on the computer the graph seems to be flipped upside down from a normal cos-1 graph.
Any help?
Again I am stuck (what's new?):
I need to graph y = cos(2x - pi/4)
I know it's y=cos2x shifted to the right by pi/8. I just don't know the technique to graph it.
Also i know amplitude is 1; period (T) = pi ... yet i still can't graph it.
EDIT:While I'm here, I might also ask how...
Could anybody tell me how to approach these types of questions? (trig functions):
e.g. lim(x --> 0) sin3x/5x or lim(x-->0) 1/x sin(x/2pi)
Tbh i don't know how to do any of these types so could someone explain it to me?(i went through the textbook, but got confused).
Rather than create my own thread, I'll just post a question here:
A circle has chord of 25mm with an angle of pi/6 subtended at the centre. Find, to 1 decimal place, the length of the arc cut off by the chord.
Ah ok, this stuff is slowly coming back to me (it's been a while since i've worked with inequalities).
So if you get a harder inequality (that's harder to sketch), you would just test points then?
This is an incredibly stupid question from me (which didn't warrant the creation of a new thread just for it - so i'll stick it here). Please help me solve this simple inequality:
x2-1 > 0 (> = greater than or equal to in this case).
What i do:
x2 > 1
x > plus/minus 1
therefore x > 1 , and...
Ok ty for that. I think i get it.
For q 2, i am a little unsure how to find the domain. This is what i do:
-1 < sqrt(1-x2) < 1 (where "<" means less than or equal to)
I solve that but something is wrong because i get the domain to be:
0 < x < 0 , which is obviously wrong...