StudyHard123
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What is the domain and range of the function
y = pi/6 - 2sin^-1 x^2
Sketch that curve.
y = pi/6 - 2sin^-1 x^2
Sketch that curve.
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Curious inverse trig. question (1 Viewer)
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Bennett Hadley
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https://www.wolframalpha.com/input/?i=y+=+pi/6+-+2sin^-1+x^2
Check this out
Check this out
That's a tricky question. Nevertheless, I'll have a go at it.
First of, I think it's going to be easier to determine domain and range from looking at the graph rather than using algebra. To draw the graph, first consider y=sin^-1x. The domain and range of sin^1x are -1<x<1 and -pi/2<y<pi/2 respectively.
Now consider y=sin^-1(x^2). This essentially means that for all x, y must be a positive value now. The intersection of the graphs will be at (1,1), and y=sin^-1(x^2) will curve below y=sin^1x for 0<x<1 (since any number between 0 and 1 squared becomes an even smaller number). Since it is an odd function, you can simply reflect the y=sin^-1(x^2) about the x-axis, and that completes the graph y=sin^-1(x^2). Note that the domain has not changed, and range has changed to 0<y<pi/2. The graph should look similar to a parabola.
2sin^-1(x^2) means that the graph's amplitude doubles, so the new range will be 0<y<pi, but x still has to be between -1 and 1. Therefore at this point you might be able to tell that the lines x=-1 and x=1 start behaving like asymptotes.
-2sin^-1(x^2) is just your previous graph flipped about the x-axis, which should look like an concave down parabola. The range is now -pi<y<0
Lastly, the pi/6 means the whole graph is shifted up by pi/6 units. The domain is still -1<x<1 whereas the new range is now:
-pi + pi/6 < y < 0 + pi/6
-5pi/6 < y < pi/6
The graph should then just look like a concave down parabola with y-intercept pi/6 with the domain and range listed above.
Hopefully I did this all correct. It probably will be confusing to look at firstly but if you write down everything and graph the initial functions we used to get to the final function on a piece of paper, it might make a bit more sense.
First of, I think it's going to be easier to determine domain and range from looking at the graph rather than using algebra. To draw the graph, first consider y=sin^-1x. The domain and range of sin^1x are -1<x<1 and -pi/2<y<pi/2 respectively.
Now consider y=sin^-1(x^2). This essentially means that for all x, y must be a positive value now. The intersection of the graphs will be at (1,1), and y=sin^-1(x^2) will curve below y=sin^1x for 0<x<1 (since any number between 0 and 1 squared becomes an even smaller number). Since it is an odd function, you can simply reflect the y=sin^-1(x^2) about the x-axis, and that completes the graph y=sin^-1(x^2). Note that the domain has not changed, and range has changed to 0<y<pi/2. The graph should look similar to a parabola.
2sin^-1(x^2) means that the graph's amplitude doubles, so the new range will be 0<y<pi, but x still has to be between -1 and 1. Therefore at this point you might be able to tell that the lines x=-1 and x=1 start behaving like asymptotes.
-2sin^-1(x^2) is just your previous graph flipped about the x-axis, which should look like an concave down parabola. The range is now -pi<y<0
Lastly, the pi/6 means the whole graph is shifted up by pi/6 units. The domain is still -1<x<1 whereas the new range is now:
-pi + pi/6 < y < 0 + pi/6
-5pi/6 < y < pi/6
The graph should then just look like a concave down parabola with y-intercept pi/6 with the domain and range listed above.
Hopefully I did this all correct. It probably will be confusing to look at firstly but if you write down everything and graph the initial functions we used to get to the final function on a piece of paper, it might make a bit more sense.
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Sorry, not sure why a big part of my solution gets cut out after "will curve below y=sin^-1x for 0", tried editing it but it's not showing up, for me at least. Here's what's meant to be after it:
Last edited: