Finding inverse, given function and domain restrictions

[dnymeyer]

Please help me understand how given domain restrictions changes how I solve for an inverse function?

I am given
f(x) = x² + 1, (-infinity, 0]

I understand that to find the inverse, I will interchange x/y variables and solve for y.

x = y² + 1
solve to:
y = +(square root of x - 1), -(square root of x-1)

What I do not understand, is how the domain restriction changes how I solve this equation? (if it does?)

 

[pka]

Please help me understand how given domain restrictions changes how I solve for an inverse function?
I am given f(x) = x² + 1, (-infinity, 0]
I understand that to find the inverse, I will interchange x/y variables and solve for y.
x = y² + 1
solve to:
y = +(square root of x - 1), -(square root of x-1)
What I do not understand, is how the domain restriction changes how I solve this equation? (if it does?)

Note that \(\displaystyle f(x)>0\) but \(\displaystyle x<0\), so the inverse output is negative.
\(\displaystyle y=-\sqrt{x-1}\) where \(\displaystyle x\in [1,\infty)\).

 

[mmm4444bot]

… I am given

f(x) = x² + 1, (-infinity, 0]

I've assumed that half-open set is a restricted domain of function f. In other words, the natural domain of x^2+1 has been restricted; you're concerned only with the left-half of f's parabola and the corresponding half-parabola for f-inverse.

… What I do not understand, is how the domain restriction changes …

The restricted domain of the original function becomes the restricted range of the inverse function.

The restricted range of the original function becomes the restricted domain of the inverse function.

That is, the domain and range switch places. If you sketch f(x) and its inverse, the restricted domains and ranges for each function will be clear. Point (0,1) on f's graph becomes point (1,0) on the graph of f-inverse, etc. :cool:

 

[pka]

Please help me understand how given domain restrictions changes how I solve for an inverse function?
I am given f(x) = x² + 1, (-infinity, 0]

I've assumed that open set is a restricted domain of function f. In other words, the natural domain of x^2+1 has been restricted; you're concerned only with the left-half of the parabola.

The set \(\displaystyle (-\infty,0]\) is decidedly not an open set. Geometrically that set is a ray.
The set \(\displaystyle (-\infty,0)\) is an open set known as a half line.

 

[Steven G]

Please help me understand how given domain restrictions changes how I solve for an inverse function?

I am given
f(x) = x² + 1, (-infinity, 0]

I understand that to find the inverse, I will interchange x/y variables and solve for y.

x = y² + 1
solve to:
y = +(square root of x - 1), -(square root of x-1)

What I do not understand, is how the domain restriction changes how I solve this equation? (if it does?)

I am given
f(x) = x² + 1, (-infinity, 0] So x is in (-infinity, 0] for f(x). In the inverse, as YOU stated, x and y are interchanged. So in the inverse y is in (-infinity, 0]. That is y is not positive. Not that the square root of (x-1) is never negative. Go from here.

 

[dnymeyer]

Okay, thanks! So what I am understanding is, the steps to algebraically solve this doesn't change, except that when I graph it, I am only concerned with the portion of the graph which y < 0

 

[Steven G]

Okay, thanks! So what I am understanding is, the steps to algebraically solve this doesn't change, except that when I graph it, I am only concerned with the portion of the graph which y < 0

Yes, so what is the equation of this graph (the inverse)???

 

[pka]

Okay, thanks! So what I am understanding is, the steps to algebraically solve this doesn't change, except that when I graph it, I am only concerned with the portion of the graph which y < 0

No \(\displaystyle y\not<0\). Because The set \(\displaystyle f(0)=0^2+1=1\) then the preimage \(\displaystyle f^{-1}(1)=0\) so we use \(\displaystyle [1,\infty)\) as the domain of the inverse. See my reply above.

 

[Steven G]

No \(\displaystyle y\not<0\). Because The set \(\displaystyle f(0)=0^2+1=1\) then the preimage \(\displaystyle f^{-1}(1)=0\) so we use \(\displaystyle [1,\infty)\) as the domain of the inverse. See my reply above.

Sorry, but you did not answer my question. What is the equation of the inverse along with any domain restrictions. Are you sure that y is never negative? Does your graph match that??

 

[Steven G]

Sorry, but you did not answer my question. What is the equation of the inverse along with any domain restrictions. Are you sure that y is never negative? Does your graph match that??

Remember that any statement you can say about x (or y) in f(x), you can say that same statement about y (or x) in f ^-1(x).

 

[pka]

Remember that any statement you can say about x (or y) in f(x), you can say that same statement about y (or x) in f ^-1(x).

Note that \(\displaystyle f(x)>0\) but \(\displaystyle x<0\), so the inverse output is negative.
\(\displaystyle y=-\sqrt{x-1}\) where \(\displaystyle x\in [1,\infty)\).

Do you bother to read the replies?
The initial set of the function \(\displaystyle f(x)=x^2+1\) is given as \(\displaystyle (-\infty,0]\) which means that the finial set is \(\displaystyle [1,\infty)\).
So yes \(\displaystyle y\ge 1\) so the domain of \(\displaystyle f^{-1}(x)\) is \(\displaystyle [1,\infty)\)

 

[mmm4444bot]

The set \(\displaystyle (-\infty,0]\) is decidedly not an open set.

You're right; I was thinking 'half-open' set. I've fixed my typo, thanks. :cool:

 

[Steven G]

You're right; I was thinking 'half-open' set. I've fixed my typo, thanks. :cool:

I was taught they were called clopen sets