Inverse Trig (1 Viewer)

[Petyo]

How would you find the range of y= xcos^-1 x ??

 

[Drongoski]

Edit

Solution posted earlier was massively wrong.

 

[Petyo]

The domain, or x should be from -1 to 1.
btw the answer in the text book (the range) is from (-pi) to less than (pi/2).

 

[Drongoski]

The domain, or x should be from -1 to 1.
btw the answer in the text book (the range) is from (-pi) to less than (pi/2).

Of course. How silly of me. Values of cos can only vary from -1 to 1.

 

[lolokay]

the lower limit is easy to find (just use x=-1).

the upper limit is obv wrong. it can't rly be found, afaik.

 

[bored of sc]

How would you find the range of y= xcos^-1 x ??

The domain of inverse cosine is -1 < x < 1.

When you sub in x = -1, y = -pi
When you sub in x = 0 or 1, y = 0

But when you sub in say x = 1/2, y = pi/6

So it's confusing.

 

[Ostentatious]

Try using Graphmatica to graph it. It's totally trippy.

 

[Trebla]

Since the domain is -1 ≤ x ≤ 1, what you would typically do is substitute the end points of the domain to find the range. However, this assumes that the function you're dealing with is monotonic increasing or decreasing (i.e. has no turning points).

In this particular question, there happens to be a maximum turning point. The problem is that you can't really find the roots of this equation:



to determine the maximum turning point. So the exact upper limit is hard to find.

 

[gurmies]

Since the domain is -1 ≤ x ≤ 1, what you would typically do is substitute the end points of the domain to find the range. However, this assumes that the function you're dealing with is monotonic increasing or decreasing (i.e. has no turning points).

In this particular question, there happens to be a maximum turning point. The problem is that you can't really find the roots of this equation:



to determine the maximum turning point. So the exact upper limit is hard to find.

What would happen if you were to approximate the root of that equation and keep approximating and subbing in, and observe what it approaches? Should in theory approach pi/2?

 

[lolokay]

it's not gonna really get anywhere near pi/2

max: x=0.6522, y=0.5611

 

[Petyo]

it's not gonna really get anywhere near pi/2

max: x=0.6522, y=0.5611

Where did you get these figures from?
Btw In class my teacher 'solved' this question by making a table of values. She has always been so unreliable...

 

[lolokay]

Where did you get these figures from?

graphmatica

 

[zazzy1234]

How would you find the range of y= xcos^-1 x ??

the range would be the inverse of cosx
cosx:
domain: 0≤ x≤ pi
range: -1≤y≤1

therefore the inverse is:
xcos^-1 is
domain:-1≤x≤1
range:0≤y≤pi

 

[gurmies]

the range would be the inverse of cosx
cosx:
domain: 0≤ x≤ pi
range: -1≤y≤1

therefore the inverse is:
xcos^-1 is
domain:-1≤x≤1
range:0≤y≤pi

?

 

[Trebla]

You haven't considered the effect of multiplication by x to inverse cosine on the range...

 

[gurmies]

 

[zazzy1234]

?

what's hard 2 understand lol

u just swap the range and domain around when it comes 2 inverse =p

 

[gurmies]

what's hard 2 understand lol

u just swap the range and domain around when it comes 2 inverse =p

Look at the graph above, the x dilates it.

 

[zazzy1234]

^ooooooooooooo i see

 

[gurmies]

^ooooooooooooo i see

=)