inverse trig range (1 Viewer)

[SaHbEeWaH]

hi
can somebody show me how to find the range of
y = x.cos^-1x

thanks

 

[Slidey]

It's the same as for arccos(x).

y=x has all real, but arccos(x) has -1 to 1, so all up it is -1 to 1.

 

[SaHbEeWaH]

are you sure?
i thought the domain was -1 ≤ x ≤ 1
it says here the range is -pi ≤ y < pi/2, but i don't know how to get that..

 

[richz]

isnt this a 4u question?????

 

[SaHbEeWaH]

i dno, should be 3u..
it comes from ext1 maths in focus

 

[Slidey]

are you sure?
i thought the domain was -1 ≤ x ≤ 1
it says here the range is -pi ≤ y < pi/2, but i don't know how to get that..

I misread it, sorry. I thought you wanted domain.

Well you could use calculus:
f(-1)=-pi
f(1)=0
f'(x)=x.(arccos(x))'+arccos(x)
I don't know inverse trig, but solve that for f'(x)=0 and then build a range.

 

[Will Hunting]

Nah, not 4U.

If cos^-1x = u
Then cos u = x, -1 &le; x &le; 1

There's your domain. Because it's restricted, all you need to do now for the range is find the minimum and maximum values of the function for -1 &le; x &le; 1.

Start with the endpoints:

y = -1 x cos^-1-1 = -pi
y = 1 x cos^-11 = 0

For the maximum/minimum values, differentiate:

First, put y = cosu.u
Then,

y' = cosu.u' - u.sinu = -cosu/sinu - u.sinu = 0 (From dx/du = -sinu, du/dx = -1/sinu)

From this, it can be seen that the graph, y = xcos^-1x has discontinuity at u = pi/2 (i.e. x = 0, y = pi/2), since y' is undefined for that value of u.

We know, by substitution, that the graph exists for x = 0 (in the point (0,0)). But y' is undefined for x = 0, &there4; (0,0) is a critical point.

However, testing for small values of x ---> 0⁺, the graph is seen to approach pi/2. Hence, we can infer that y = xcos^-1x has a critical point at (0,0) AND a point of discontinuity at (0, pi/2), or has a filled in circle at (0,0) and an empty circle at (0, pi/2).

&there4; the range is -pi &le; y < pi/2

---

Alternatively, if you want to throw 4U into the mix, just graph y = cos^-1x and y = x on the same set of axes and use a multiplication of ordinates to inspect for maximum and minimum values

 

[haboozin]

This question in MIF had the correct answer, but not all of the range questions in that section of the chapter have correct answers. There is 1 wrong range limit asnwer and 1 wrong graphing answer (I think its like the last one with the tan's) MIF forgets that the x-axis is continueous..

 

[Pace_T]

Hey will hunting (or someone else), could you please explain your post if you don't mind
"We know, by substitution, that the graph exists for x = 0 (in the point (0,0)). But y' is undefined for x = 0, ∴ (0,0) is a critical point."

critical point has 2 y values, 1 x value?


"However, testing for small values of x ---> 0+, the graph is seen to approach pi/2."

say u put in 0.001 as x, then y = 0.001*cos-1(0.001)
= 0.008 or somewhere around there
how come mine does not approach pi/2?

lol thanks for ur time
greatly appreciated.

 

[Slidey]

y'=arccos(x)-x/sqrt(1-x^2)=0
y'(0.5)=0.47
y'(0.75)=-0.41
Let's test x=0.65
y'(0.65)=0.008
Hum hum hum.
y(0.65)=0.56

My point is that the max is 0.56 or so. Unless there is some very strange discontinuity I am missing, pi/2~=1.6 is far too high.

Oh, and (0,0) isn't a critical point.
y'(0)=pi/2

It'll be interesting to see Will's justification. I'm leaning towards the book being wrong.

 

[Pace_T]

how did u know that u have to convert the 1.6 into the pi stuff
like, i thought that was the degrees in 'y(0.65)=0.56'
I must be really dumb lol

 

[Slidey]

Always use radians for graphing.

pi is just 3.14 et cetera. Half pi is roughly 1.6 I was just demonstrating that the maximum y value is actually a little over pi/5. I don't know how the book or Will got pi/2.

 

[Pace_T]

hey so was will hunting's answer right?
i just wanna know cause im 50% sure my teacher will put that qu in our test that is on tues
thanks for any help

 

[Slidey]

As far as I can tell, no, he's not.

1) The critical points are actually at x=1 and -1.
2) The maximum value does not appear to be pi/2 as the answer says, due to the derivative.
3) The derivative is continuous on the interval (-1,1)
4) The function itself is continuous on the interval (-1,1)
5) I graphed it by multiplication of functions and I see no anomalies - just a maximum roughly at the point (0.65,0.56)

Graphmatica would agree: