inverse of a function

[Dorian Gray]

Hello Everyone,

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.

Thank you

 

[burakaltr]

Hello Everyone,

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.

Thank you

For starters, multiply and divide the right part with the numerator.

 

[Dorian Gray]

?

@burakaltr Thanks for the info. I multiplied it out and got this. However, I am not sure what I am supposed to do with this.


@JeffM Thank you for your advice; however, I still don't quite understand what you are saying.

 

[tkhunny]

Why on Earth would you do that?

#1 thing to do. Look at it. Domain? x >= 0 Range? y > -1 I cannot sufficiently stress the importance of this. It will save you at all times.

y * (1+sqrt(x)) = 1-sqrt(x)

y + y*sqrt(x) + sqrt(x) = 1

sqrt(x)*(y+1) = 1-y

sqrt(x) = (1-y)/(1+y)

How close are we? Are we getting anywhere?

Now, why did we cosider the Domain and Range before we got started with the algebraic manipulation?

 

[pka]

@JeffM Thank you for your advice; however, I still don't quite understand what you are saying.

Yours is the usual way. So stick to it.
If you have \(\displaystyle x=\dfrac{1-\sqrt{y}}{1+\sqrt{y}}\) then \(\displaystyle x+x\sqrt{y}=1-\sqrt{y}\)

so \(\displaystyle \sqrt{y}=\dfrac{1-x}{1+x}\text{ or }y=\left(\dfrac{1-x}{1+x}\right)^2\).

 

[Dorian Gray]

thank you

Thank you pka for your advice and insight. It was certainly appreciated.


And thank you to everyone else who helped with the problem.

 

[lookagain]

Why on Earth would you do that?

#1 thing to do. Look at it. Domain? x >= 0 Range? y > -1 Actually, the range is

\(\displaystyle -1 < y \le 1 \ \ \ or \)

\(\displaystyle (-1, 1]\)

And numerically, the domain of the inverse function is the range of the given function.

So, the domain of the inverse is

\(\displaystyle -1 < x \le 1 \ \ \ or\)

\(\displaystyle (-1, 1]\)


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 \(\displaystyle f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}} \ \ is\)

this entire last line:


\(\displaystyle f^{-1}(x) \ = \ \bigg(\dfrac{1 - x}{1 + x}\bigg)^2, \ \ -1 < x \le 1\)





\(\displaystyle \text{I know about this, because I gave it as a challenge problem }\)

\(\displaystyle \text{when I first started posting in this forum.}\)

 

[Dorian Gray]

thank you

Thank you for pointing that out Lookagain.


Thanks again to everyone who helped with the problem

 

[pka]

(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 \(\displaystyle y(x)=\dfrac{1-\sqrt{x}}{1+\sqrt{x}}~\&~f(x)=\left(\dfrac{1-x}{1+x}\right)^2\) do have an interesting relationship. As noted in the above quote they are inverses of each other on \(\displaystyle [0,1]\).
On the domain of \(\displaystyle y:~[0,\infty)\) the function \(\displaystyle y\) is one-to-one so an inverse exits.
Now, the function \(\displaystyle f\) is the left-inverse of \(\displaystyle y\) on the domain \(\displaystyle [0,\infty)\)
That means for all \(\displaystyle x\in[0,\infty)\) we have \(\displaystyle f\circ y(x)=f(y(x))=x\).
But for \(\displaystyle x\in(1,\infty)\) we have \(\displaystyle y\circ f(x)=y(f(x))\ne x\).

 

[Deleted member 4993]

Then, the correct inverse function for \(\displaystyle f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}} \ \ is\)

this entire last line:


\(\displaystyle f^{-1}(x) \ = \ \bigg(\dfrac{1 - x}{1 + x}\bigg)^2, \ \ -1 < x \le 1\)

Since f(x) does not exist for x<0 in real domain - what would be the meaning of f^-1(x) for x<0

 

[lookagain]

Then, the correct inverse function for \(\displaystyle f(x) \ = \ \dfrac{1 - \sqrt{x}}{1 + \sqrt{x}} \ \ is\)

this entire last line:


\(\displaystyle f^{-1}(x) \ = \ \bigg(\dfrac{1 - x}{1 + x}\bigg)^2, \ \ -1 < x \le 1\)

Since f(x) does not exist for x<0 in real domain - \(\displaystyle > > \) what would be the meaning of f^-1(x) for x<0 \(\displaystyle < < \)

Subhotosh Khan,

possibly there is temporary confusion about the domains and ranges back and forth
between f and the inverse of f.


I stated that the domain of the inverse of f is \(\displaystyle -1 < x \le 1.\)


You asked (highlighted in the quote box by me) about what would be the meaning of \(\displaystyle f^{-1}(x) \ for \ x < 0.\)


That portion of the domain of \(\displaystyle f^{-1}\) would then be -1 < x < 0.


Suppose, for example, you take x = -1/2 for an x-value to be substituted in for x in \(\displaystyle f^{-1}(x).\)


Evaluating \(\displaystyle f^{-1}(-1/2)\) gives 9.


This value of 9, in the range of f(x) for x = -1/2, corresponds to x = 9 in the domain of f(x).


An x-value can be equal to 9 in the domain of f, with the value of f
(as part of its range) equal to -1/2.