What is the inverse of #h(x)=-4/x^2#?
The furthest I got on this question was #y=sqrt(-1/4x#
But I don't think that's right
The furthest I got on this question was
But I don't think that's right
1 Answer
Oct 23, 2016
Do you want + or -?
Explanation:
Exchange x and y.
Multiply by
Divide by x, then take the root.