Expert: Paul Klarreich - 5/26/2007

Question
QUESTION: y=x arccosx :determine the domain and range domain is o.k range i cannot solve without a graphical tool solution

ANSWER: Questioner:   bruce
Category:  Advanced Math
Private:  No
 
Subject:  inverse trig range
Question:  y=x arccosx :determine the domain and range domain is o.k range i cannot solve without a graphical tool solution

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Hi, Bruce,

For  y = arccos x, the domain is [-1,1] and the range is generally taken to be  [0,pi]   But if that is multiplied by x, to get your function:

y = f(x) = x arccos x

now you have something different.  You can make these observations:

f(-1) = -1(arccos(-1)) = -1(pi) = -pi
f(1)  = 1(arccos(1)) = 1(0) = 0.
f(0) = 0
But if  x is in [0,1], arccos(x) is positive and so is x, and therefore we have some positive function values.  Therefore we have to investigate the maximum values, and have to find critical points:

         x
f'(x) = arccos(x) + -------------
         sqrt(1 - x^2)  

That's going to be a bit messy, but the maximum will be somewhere about f(x) = 0.4.  The minimum is f(-1) = -pi, so that is your range:  [-pi, 0.4]
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Now a request:

Did you mean to write:

For  y = x arccos x, determine the domain and range[PERIOD].

The domain is OK [PERIOD].

The range I cannot determine without a graphical tool[PERIOD]

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In other words, please use those extra buttons on the keyboard -- shift keys and punctuation marks.  You are not sending me an instant message; there is no rush to get it sent, so you can be careful about capitalization and punctuation.  Then your question makes more sense and you are likely to get a better answer.

Would you like me to send an answer looking like this?

for  y = arccos x the domain is [-1,1] and the range is generally taken to be  [0,pi]  but if that is multiplied by x, to get your function y = f(x) = x arccos x now you have something different you can make these observations:

f(-1) = -1(arccos(-1)) = -1(pi) = -pi  f(1)  = 1(arccos(1)) = 1(0) = 0  f(0) = 0   but if  x is in [0,1] arccos(x) is positive and so is x and therefore we have some positive function values therefore we have to investigate the maximum values, and have to find critical points:

         x
f'(x) = arccos(x) + -------------
         sqrt(1 - x^2)  

thats going to be a bit messy but the maximum will be somewhere about f(x) = 0.4 the minimum is f(-1) = -pi so that is your range


---------- FOLLOW-UP ----------

QUESTION: Tanks for your prompt reply.

The range of the f(x)= x. arccosx is still troubling me.

I have found it through an interative method(trial and error)but was hoping for a more accurate result.

Is there a way to solve the derivative of this function to find the upper bound of the range?

If not what approximation method would best be suited to find an approximate value?

With thanks Bruce Hill

Answer
Hi, Bruce,

You are trying to solve this equation:

         x
f'(x) = arccos(x) + ------------ = 0
         sqrt(1 - x^2)

I can't think of any exact way so solve it, so you will have to use some iterative method.

I think that its derivative (that would be f''(x)) is not close to zero near the solution point.  That means you could use Newton's method.  Calling that function  g(x), for simplicity,
         x
g(x) = arccos(x) + ------------
         sqrt(1 - x^2)

you start by finding  g'(x).  Messy, but not impossible.  Next choose a starting value of x, perhaps  x1 = 0.5.

Now Newton's method says:

x[n+1] = x[n] -  g( x[n] )/g'( x[n] )

I.e.  plug in that x1 and you get x2.  Plug in x2 and you get x3, etc.  [An Excel spreadsheet is a good way to set it up.]

If the original value is well chosen, and the function does not misbehave (that usually means the derivative is zero nearby) then Newton's method converges quite rapidly.