Question 1208035
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I'll provide a supplement to the other tutor's work.
Specifically I'll show another way of solving {{{1-x^2 >= 0}}}


{{{1-x^2 >= 0}}}


{{{1>=x^2}}}


{{{x^2<=1}}}


{{{sqrt(x^2)<=sqrt(1)}}} Since the square root function is an increasing function, applying it to both sides does <u>not</u> flip the inequality sign.


{{{abs(x)<=1}}}


{{{-1<=x<=1}}} is the domain


The rule I used was |x| < k is the same as -k < x < k for some positive real number k.
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