Sketching and inverse sine graph from a GIVEN f(x) (1 Viewer)

[blackops23]

Hi guys just need a bit of help with this one, here's how it goes:

f(x) is an odd function, as x--> +/- inf, f(x) --> 0, there is a MAXIMUM at (1,1) and a MINIMUM at (-1,-1) and a point of inflexion at (0,0).

I had to sketch g(x) = inv.sin(f(x)).
Here's what I did:
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Range of f(x): -1<= f(x) <= 1
So: inv.sin(-1) <= inv.sin(f(x)) <= inv.sin(1)

-pi/2 <= inv.sin f(x) <= pi/2

Domain of inv.sin (x) is: -1< (x) < 1
So for y= inv.sin f(x)

DOMAIN: -1<= f(x) <=1, therefore ALL REAL X.

Now f(x)=0 when x=0, g(x) = 0, therefore g(x) passes through (0,0)
f(x) = 1 when x =1, g(x)= pi/2, therefore (1, pi/2)
Similarly, g(x) passes through (-1, -pi/2)

Also, as x--> +/- inf, f(x) --> 0, g(x)=f(x) as x--> +/- inf.
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And so I sketched it with nice curvy turning points at (1, pi/2) and (-1, pi/2) with the graph --> 0, as x--> +/- inf. So it looked very similar to f(x)

Then I looked at the answers in my textbook, and to my surprise, the turning points weren't curvy at all, rather they were sharp and had a corner... So obviously the gradient at x=+/- 1 was INDETERMINABLE.

Say I'm in an exam, how the heck am I supposed to know when f'(x) is INDETERMINABLE? What would you have done in a test/exam to know that there were two sharp corners on the turning points??

 

[Bored Of Fail 2]

g(x) = inv sin ( f(x) )
u= f(x)

g(x) = inv sin ( u )

g'(x) = f ' (x) / sqrt ( 1- (f(x)) ^2 )

which is not defined when f(x) = +- 1 ( because you divide by zero ) and f(x) = +- 1 when x= +1 or x=-1 respectively

 

[blackops23]

Quick question: What is the difference between UNDEFINED (1 divided by 0) and INDETERMINABLE ( 0/0 or infinity/infinity, or 0*infinity, or infinity MINUS infinity)???

Thanks

 

[Bored Of Fail 2]

Quick question: What is the difference between UNDEFINED (1 divided by 0) and INDETERMINABLE ( 0/0 or infinity/infinity, or 0*infinity, or infinity MINUS infinity)???

Thanks

to be honest I dont know, I think technically it is all called "indeterminable" , but they are not going to ask that in an exam

 

[blackops23]

to be honest I dont know, I think technically it is all called "indeterminable" , but they are not going to ask that in an exam

Ok, textbook says, if y' = [f'(x)/($$$)], and $$$= 0 , but f'(x) doesn't equal zero, then y' --> infinity, i.e. vertical tangent. Looks like in the, y=inv.sin(f(x), f'(x) happened to equal zero... I guess in an exam you would just guess that f'(x) equals zero, because you can't have a vertical tangent on a minima/maxima, so it's got to be indeterminable. Wouldn't you agree?

 

[Bored Of Fail 2]

lol im so lost

 

[blackops23]

lol im so lost

lol, that's crazy talk, youre a crap load smarter than me... mebbe i just can't explain very well

 

[jyu]

Quick question: What is the difference between UNDEFINED (1 divided by 0) and INDETERMINABLE ( 0/0 or infinity/infinity, or 0*infinity, or infinity MINUS infinity)???

Thanks

'undefined' means the limit does not exist.
'indeterminable' means the limit may/may not exist.

 

[blackops23]

'undefined' means the limit does not exist.
'indeterminable' means the limit may/may not exist.

cool thanks, I looked at the graph of a derivative that had a point make it equal 0/0, and it was really weird, the graph just jumped up to a random point on the same vertical line..