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inverse trig range (1 Viewer)
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Slidey
But pieces of what?
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I misread it, sorry. I thought you wanted domain.SaHbEeWaH said:
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
Member
Nah, not 4U.
If cos-1x = u
Then cos u = x, -1 ≤ x ≤ 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 ≤ x ≤ 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, ∴ (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).
∴ the range is -pi ≤ 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
If cos-1x = u
Then cos u = x, -1 ≤ x ≤ 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 ≤ x ≤ 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, ∴ (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).
∴ the range is -pi ≤ 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
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haboozin
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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
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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.
"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
But pieces of what?
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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.
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.
Last edited:
Slidey
But pieces of what?
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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:
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: