Since y = arccos((1-x^2)/2), y has an imaginary component if
(1-x^2)/2 < -1
1-x^2 < -2
3 < x^2
y is real if -sqrt(3) <= x <= sqrt(3)
x = -1 falls within this range and is the absolute minimum of x-(1-x^2)/2 which is -1, the vertex of a concave up parabola.
At x=sqrt(3), x-(1-x^2)/2 reaches its maximum within the allotted domain, 1 + sqrt(3).
-1 to 1+sqrt(3) is the range.
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Posted by Charlie
on 2021-09-20 10:21:05 |
solution
