I am having difficulties with the following problem. I am trying to find the inverse function of this
I've replaced the y with the x, but should I get rid of the fraction first? Should I square it first? Any comments, help, suggestions, or advice is appreciated.
I am having difficulties with the following problem. I am trying to find the inverse function of this View attachment 1617
I've replaced the y with the x, but should I get rid of the fraction first? Should I square it first? Any comments, help, suggestions, or advice is appreciated.
And numerically, the domain of the inverse function is the range of the given function.
So, the domain of the inverse is
−1<x≤1or
(−1,1]
And that function that pka typed for the last line is not a one-to-one function,
so it has to be (properly) restricted with the domain that I typed just above in
order to be a one-to-one function.
Then, the correct inverse function for f(x)=1+x1−xis
this entire last line:
f−1(x)=(1+x1−x)2,−1<x≤1
I know about this, because I gave it as a challenge problem
(The) function that pka typed for the last line is not a one-to-one function, so it has to be (properly) restricted with the domain that I typed just above in order to be a one-to-one function.
The functions y(x)=1+x1−x&f(x)=(1+x1−x)2 do have an interesting relationship. As noted in the above quote they are inverses of each other on [0,1].
On the domain of y:[0,∞) the function y is one-to-one so an inverse exits.
Now, the function f is the left-inverse of y on the domain [0,∞)
That means for all x∈[0,∞) we have f∘y(x)=f(y(x))=x.
But for x∈(1,∞) we have y∘f(x)=y(f(x))=x.
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