document.write( "Question 1075749: q = sin-1(x)+cos-1(x)-tan-1(x) for the domain x ≥ 0,
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Algebra.Com's Answer #690401 by Edwin McCravy(20056)\"\" \"About 
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document.write( "q = sin-1(x)+cos-1(x)-tan-1(x) for the domain x ≥ 0,\r\n" );
document.write( "what is the range?\r\n" );
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document.write( "Because of the inverse sine and cosine functions, \r\n" );
document.write( "the domain of q is 0 ≤ x ≤ 1, \r\n" );
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document.write( "We find the derivative of q\r\n" );
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document.write( "\"d%28theta%29%2Fdx\"\"%22%22=%22%22\"\"1%2Fsqrt%281-x%5E2%29-1%2Fsqrt%281-x%5E2%29-1%2F%281%2Bx%5E2%29\"\r\n" );
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document.write( "The first two terms cancel and we have\r\n" );
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document.write( "\"d%28theta%29%2Fdx\"\"%22%22=%22%22\"\"-1%2F%281%2Bx%5E2%29\"\r\n" );
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document.write( "Since the derivative of q is negative, the function q\r\n" );
document.write( "decreases everywhere on the open part of its domain,\r\n" );
document.write( "which is 0 < x < 1.\r\n" );
document.write( "[q is not differentiable at the endpoints 0 and 1, although \r\n" );
document.write( "q is defined at those endpoints]\r\n" );
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document.write( "We evaluate q at the endpoints of the \r\n" );
document.write( "domain of q : \r\n" );
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document.write( "Substitute x = 0\r\n" );
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document.write( "q = sin-1(0) + cos-1(0) - tan-1(0) = 0° + 90° - 0° = 90°\r\n" );
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document.write( "Substitute x = 1\r\n" );
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document.write( "q = sin-1(1) + cos-1(1) - tan-1(1) = 90° + 0° - 45° = 45°\r\n" );
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document.write( "Thus the range for q for the domain\r\n" );
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document.write( "0 ≤ x ≤ 1\r\n" );
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document.write( "is\r\n" );
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document.write( "45° ≤ q ≤ 90°\r\n" );
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document.write( "Edwin
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