Question 1203320
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(a) {{{y=xsin^(-1)(x^2)}}}<br>
domain...<br>
x^2 is 0 or positive, and the argument for inverse sine must be between -1 and 1. So the domain for {{{y=sin^(-1)(x^2)}}} is from -1 to 1; and since x is between 0 and 1, the domain for {{{y=xsin^(-1)(x^2)}}}, is still from -1 to 1.<br>
ANSWER: -1 to 1 (your answer is correct)<br>
range...<br>
{{{y=sin^(-1)(x^2)}}} has its maximum value when x^2=1, which occurs at both -1 and 1, the boundaries of the domain. At x=-1 or x=1, where x^2 is 1, inverse sine is pi/2.  So at x=-1 the value of {{{y=xsin^(-1)(x^2)}}} is -pi/2, and at x=1 the value is pi/2.<br>
ANSWER: -pi/2 to pi/2 (your answer is correct)<br>
(b) {{{y=tan^-1 (sqrt(1-x^2))}}}<br>
domain...<br>
The argument for square root must be non-negative, so again x must be between -1 and 1.<br>
ANSWER: -1 to 1 (your answer is correct)<br>
range...<br>
At both x=-1 and x=1, 1-x^2 is 0, and inverse tangent of 0 is 0.<br>
Between x=-1 and x=1, {{{sqrt(1-x^2)}}} is positive, with a maximum of 1 when x is 0. Inverse tangent of 1 is pi/4.<br>
ANSWER: 0 to pi/4 (your answer was 0 to pi/2)<br>