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Find the domain of the inverse of the following function:$$f(x) = kx^\frac{1}{2}$$

The inverse, using some basic algebraic rearranging, can be said to be:

$$f^{-1}(x) = \frac{k^2}{x^2}$$

As evident, the domain of this function seems to be all real numbers.

However, a property of functions states that,

The domain of $f(x)$ is equal to the range of $f^{-1}(x)$, and the domain of $f^{-1}(x)$ is equal to the range of $f(x)$

We know that the range of $f(x)$ is $x>0$;

The presumed domain of $f^{-1}(x)$ doesn't equal the range of $f(x)$. Why is the domain of $f^{-1}(x)$ said to be the range of $f(x)$ even when the inverse function doesn't supposedly need to have a restriction on its domain?

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  • $\begingroup$ Because the inverse function of $f$ depends on $f$, although it might be extended. $\endgroup$ – Friedrich Philipp Mar 20 '16 at 4:25
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    $\begingroup$ The inverse is $f^{-1}(y) = \frac{y^2}{k^2}$ $\endgroup$ – iiivooo Mar 20 '16 at 4:27
  • $\begingroup$ The range of $f(x)$ is never $(0,\infty)$. There are three things it can be, and this is none of them. $\endgroup$ – Eric Towers Mar 20 '16 at 4:59
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Firstly, the inverse needs to take any output from $f(x)$ and return the original value. As such values outside that output (the original's range) should not be part of the inverse's input (domain).

Secondly you've rearranged incorrectly and your fraction is upside down.

Thirdly, the domain of $f(x)$ is $x\geq0$ not just $x>0$.

Lastly, there are two cases you need to consider. Either $k>0$ or $k<0$. We can ignore the $k=0$ case as this does not have an inverse function.

Case $k>0$

The domain of $f(x)$ is $x\geq0$ and so we can determine that the range is $y\geq0$. Hence the domain of the inverse should be $x>0$. We have to ignore the rest of the potential domain ($x<0$, show with dotted line) as these are not possible outputs from $f(x)$.

This leads to $f^{-1}(x)=\frac{x^2}{k^2}, x\geq0$

Example with $k=1$

Example with $k=1$

Case $k<0$

Basically the same but with signs reverse. The domain of $f(x)$ is $x\geq0$ and so we can determine that the range is $y\leq0$. Hence the domain of the inverse should be $x\leq0$. We have to ignore the rest of the potential domain ($x>0$, show with dotted line) as these are not possible outputs from $f(x)$.

This leads to $f^{-1}(x)=\frac{x^2}{k^2}, x\leq0$

Example with $k=-1$

Example with $k=-1$

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  • $\begingroup$ I think the inverse function for case $k<0$ should be $f^{-1}(x)=\frac{x^2}{k^2}$ for $x\leq 0$. The image of the inverse function should be the domain of the original function, which is positive. Also you can obtain the inverse function by reflecting the original function about the line $y=x$ $\endgroup$ – lEm Mar 20 '16 at 5:45
  • $\begingroup$ Excellent correction. Will update answer. $\endgroup$ – Ian Miller Mar 20 '16 at 5:47
  • $\begingroup$ @Arjun These graphs beautifully illustrate a nice property about functions and their inverses: they show symmetry about the line $y=x$. $\endgroup$ – Mallory Mar 20 '16 at 13:54
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The inverse of the function $f(x)=kx^{\frac 12}$ isn't the entire parabola, just part of it. In order to take the inverse of a function, it has to be one-to-one. When you algebraically find the inverse, keep in mind that what you have is the positive root of $x$, not the negative.

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In order for $f$ to have an inverse, it has to be bijective. For $f(x)=kx^{1/2}$, it only takes value for $x\geq 0$. And depending on $k$, the range could non-negative real or non-positive real.

But the problem is its inverse function $f^{-1}(x)=\frac{x^2}{k^2}$ is not bijective if you take all real number as its domain. So actually the real inverse function is the restriction $f^{-1}(x)=\frac{x^2}{k^2}$ for $x\geq 0$ (if $k>0$) or for $x\leq 0$ (if $k<0$)

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  • $\begingroup$ In order for $f$ to have an inverse $f$ must be injective! The function $f$ is also not bijective and it has inverse! $\endgroup$ – iiivooo Mar 20 '16 at 4:37
  • $\begingroup$ @iiivooo it is injective because the square root of some number is specific. It is only bijective if we assign $x\geq 0$ as its domain. We also restrict the range to its image to make it surjective. So an inverse can exist $\endgroup$ – lEm Mar 20 '16 at 4:50
  • $\begingroup$ In your answer you say: "In order for $f$ to have an inverse, it has to be bijective" - From my point of view this is not mathematically exact statement! $\endgroup$ – iiivooo Mar 20 '16 at 4:58
  • $\begingroup$ @iiivooo Invertible function and bijective function are actually equivalent terms. $\endgroup$ – lEm Mar 20 '16 at 5:41
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Just to emphasize what others are saying,

the function $f$ as a function from reals to reals is actually not invertible, since it is not surjective (its range is not all of $\mathbb{R}$). In fact, it is not even defined as a function on all of $\mathbb{R}$. However, as a function from $[0,\infty)$ to the interval $[0,\infty)$ it is invertible.

It just so happens that $[0,\infty)$ is a subset of $\mathbb{R}$ and when you find an explicit formula for the inverse you need to restrict the domain of the function. That is, the actual problem you are trying to solve is this

Find an inverse for $f(x)=k\sqrt{x}$ as a function from $[0,\infty)$ to $[0,\infty)$.

The anser is: the inverse is $f^{-1}(x)=x^2/k^2$ as a function from $[0,\infty)$ to $[0,\infty)$

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  • $\begingroup$ Why do you assume $k>0$? $\endgroup$ – Eric Towers Mar 20 '16 at 5:08
  • $\begingroup$ @EricTowers Well, of course...I should say this is the case for $k>0$. The case for $k<0$ is similar and $k=0$ requires its own special treatment. $\endgroup$ – user113529 Mar 20 '16 at 5:42
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First, you don't place any restrictions on $k$. This will turn out to matter.

Second, "$f(x) = k x^{\frac{1}{2}}$" is not a function. A function presented this way is a domain and a rule/formula/expression for converting elements of the domain into elements of the image. When the domain is missing, it is normally understood to be the largest subset of whatever set makes sense given the current subject of discourse. You seem to indicate that the real numbers is that set. In this case, the domain of $f(x)$ is $x \geq 0$. (Because there are no challenges to taking square roots of zero and positive numbers.) If $k=0$, we could conceivably argue that the domain is all of the reals, but that will fail to be interesting in the next step, so we set that aside.

Now that we have a fully specified function, we can try to find its inverse. If $k=0$, we're done because $f$ is not injective (equivalently, does not pass "the horizontal line test") so does not have an inverse. (If, for some reason, the domain of $f$ had been a single point, we would be able to continue -- the domain of the inverse will be the output of $f$ at that one point.) Otherwise, if $k>0$, the range of $f$ is $[0,\infty)$ and if $k < 0$, the range of $f$ is $(-\infty,0]$. Using the fact about functions and their inverses you mention, the domain of $f^{-1}$ is either $[0,\infty)$ or $(-\infty,0]$ as $k >0$ or $k < 0$, respectively.

We could go a bit further to check the above. We compute \begin{align*} f(x) &= y = k x^{\frac{1}{2}} \text{, so} \\ f^{-1}(y) &= x = \left( \frac{y}{k} \right)^2 \text{.} \end{align*} If $k>0$, we have $y \geq 0$, and we are looking at a point on the same half of the square function that becomes the upper half of the graph of the square root function under inversion. If $k < 0$, we have $y \leq 0$, and we are looking at the same half of the graph of the square function that becomes the lower half of the square root function under inversion (as we must since $k<0$ means we are only considering that half).

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  • $\begingroup$ Since your emphasizing the effect that various values of $k$ have on the function, perhaps we could also point out the simple fact that each value of $k$ defines a different function. Hence, for each value, one has a different domain and range in general. $\endgroup$ – user113529 Mar 20 '16 at 6:53

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