Show that arcus sine is a growing function. I can explain it, but can't make a 'calculation', to prove it. Dm(Sin)=[-(2/pi);2/pi] Dm(arcsin)=[-1;1] Sin is one-to-one in the defined interval, thus Sin has an inverse function: y = f^-1(x) x = f(y) thus: y = Sin(x) arcsin(y) =x (arcsin)'(x)=0 is undefined thus f'(x) must be positive for all x which is a part of the interval of Sin. Which means arcus sine is positive. Would this be enough of an explanation ?

Question

Show that arcus sine is a growing function. I can explain it, but can't make a 'calculation', to prove it. Dm(Sin)=[-(2/pi);2/pi] Dm(arcsin)=[-1;1] Sin is one-to-one in the defined interval, thus Sin has an inverse function: y = f^-1(x) <=> x = f(y) thus: y = Sin(x) <=> arcsin(y) =x (arcsin)'(x)=0 is undefined thus f'(x) must be positive for all x which is a part of the interval of Sin. Which means arcus sine is positive. Would this be enough of an explanation ?

anonymous · Answer

For a function, f, to be growing it means that for all x_1,x_2 in the D_f, f(x_1)<=f(x_2).

anonymous · Answer

I'm guessing you already knew this.

anonymous · Answer

Ye I do ;) its a viable answer. But i was trying to see if there was a way to prove it a bit more  generally. :)

anonymous · Answer

I'm not sure how you would show it to be honest, by sin's own definition it's a strict growing function in the interval of |x|<=1 with V_f=[-Pi/2,Pi/2]

anonymous · Answer

I guess you could compare the endpoints.

anonymous · Answer

I just realized you've written what I'm walking towards, haha.

anonymous · Answer

Ye exactly. I think the question is bit misleading.

anonymous · Answer

It just says: Show that arcus sine is growing, but i've only got definitions and intervals to work with, the only way i can show it, is by explaining it,

anonymous · Answer

Look at the endpoints, since the derivative of arcsinx in the interval is never 0; you can safely assume that the function is growing.

anonymous · Answer

Yea, i'll try to do that, that sounds like a pretty good solutions :), awsome, thanks !

anonymous · Answer

Since you have that the derivative for arcsinx is equal to 1/sqrt(1-x^2), (arcsinx)'>0

anonymous · Answer

Really like your question! It's very related to what I'm taking a course in right now. Thanks!

anonymous · Answer

Yea, i wrote in the main post as: (arcsin)'(x)=0 is undefined thus f'(x) must be positive for all x which is a part of the interval of Sin. 

Since f'(x) never crosses the x-axis.

Offtopic:

Thanks mate, im a first year university student (math) so im trying to read up on a lot of stuff, mostly calculus atm. ;)

anonymous · Answer

Same here!