Find the inverse function for
1 answer:
that is correct hmmmm, yes, I think the restriction is also correct, it has to be x > 0, because, the domain on the original, has a restriction of x⩾3, otherwise the radicand turns negative.
now, the graph of that original function is a parabola sideways opening to the right, if you include both sides of the vertex of the parabola, it will not pass the vertical-line-test and not be a function, but if you restrict it to only positive roots, so it'll only be on the 1st quadrant, then it functions ok.
that means, the range of the original function, has to stick to only 0 or positive root values, now, recall that the range of the original is the domain of the inverse. Therefore, the domain of the inverse, will have to stick to only 0 or positive values, since it's bound to the same restriction as the original's range.
notice the picture below of both, the dashed section is the positive range for each, if you include it, the original will fail the vertical-line-test, and the inverse will fail the horizontal-line-test.
recall that for a function to have an inverse, they both must pass the vertical and horizontal line test, or be a one-to-one functions.
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Answer:
Lets a,b be elements of G. since G/K is abelian, then there exists k ∈ K such that ab * k = ba (because the class of ab, is equal to
, thus ab and ba are equal or you can obtain one from the other by multiplying by an element of K.
Since K is a subgroup of H, then k ∈ H. This means that you can obtain ba from ab by multiplying by an element of H, k. Thus, . Since a and b were generic elements of H, then H/G is abelian.