this is the question
$$f(x)=sin^{-1}(\sqrt{ax+b})$$
#why is squaring -1 for inequalities equals to zero T-T
median spade
queen otterBOT
nashipear
buoyant raftBOT
this is the question
$$f(x)=sin^{-1}(\sqrt{ax+b})$$
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cunning hedge
The sqrt can't be negative, that's why the -1 is irrelevant and they switched it for
$-1\le \sqrt{ax+b}\le1\to 0\le \sqrt{ax+b}\le1$
And then squred and ect
queen otterBOT
TheVinkler
median spade
The sqrt can't be negative, that's why the -1 is irrelevant and they switched it...
ohh, so for any inequality that has a sqrt it is never a negative number.
does that mean if the implied domain is negative (like inverse cos and sin) we always replace the negative number with 0?
it's just a given bc square roots can never be equals to <0?
cunning hedge
it's just a given bc square roots can never be equals to <0?
exactly