# How do you perform inversions for y = x^2 and y = x^4? Is (dx)/(dy) from the inverse 1/((dy)/(dx))?

Oct 11, 2016

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$

#### Explanation:

Assuming we are looking for the inverse functions of $f \left(x\right) = {x}^{2}$ and $g \left(x\right) = {x}^{4}$, i.e., ${f}^{- 1} \left(x\right)$ and ${g}^{- 1} \left(x\right)$ such that

$f \left({f}^{- 1} \left(x\right)\right) = {f}^{- 1} \left(f \left(x\right)\right) = x$
and
$g \left({g}^{-} 1 \left(x\right)\right) = {g}^{-} 1 \left(g \left(x\right)\right) = x$

we must consider a restricted domain. A function can only have an inverse function on a domain in which it is $1 - 1$, or else there would be ambiguity as to what to map points in the codomain back to. For example, if our initial domain is $\mathbb{R}$, should we have ${f}^{-} 1 \left(4\right) = 2$ or ${f}^{- 1} \left(4\right) = - 2$?

Again, assuming we are only considering real-valued functions, we can partition $\mathbb{R}$ into two domains, each of which results in ${x}^{2}$ and ${x}^{4}$ being $1 - 1$. We will use the notation that for $x > 0$, $\sqrt{x}$ is the principal square root of $x$, that is, the unique value $y \in {\mathbb{R}}^{+}$ such that ${y}^{2} = x$. Similarly, $\sqrt[4]{x}$ will represent the unique $y \in {\mathbb{R}}^{+}$ such that ${y}^{4} = x$.

With that, we can now consider our restricted domains for ${x}^{2}$ and ${x}^{4}$.

If we have
$f \left(x\right) : \left[0 , \infty\right) \to \left[0 , \infty\right) , f \left(x\right) = {x}^{2}$
and
$g \left(x\right) : \left[0 , \infty\right) \to \left[0 , \infty\right) , g \left(x\right) = {x}^{4}$

then we can find their inverse functions as

${f}^{-} 1 \left(x\right) : \left[0 , \infty\right) \to \left[0 , \infty\right) , {f}^{-} 1 \left(x\right) = \sqrt{x}$
and
${g}^{-} 1 \left(x\right) : \left[0 , \infty\right) \to \left[0 , \infty\right) , {g}^{-} 1 \left(x\right) = \sqrt[4]{x}$

If we have
$f \left(x\right) : \left(- \infty , 0\right] \to \left[0 , \infty\right) , f \left(x\right) = {x}^{2}$
and
$g \left(x\right) : \left(- \infty , 0\right] \to \left[0 , \infty\right) , g \left(x\right) = {x}^{4}$

then we can find their inverse functions as

${f}^{-} 1 \left(x\right) : \left[0 , \infty\right) \to \left(- \infty , 0\right] , {f}^{-} 1 \left(x\right) = - \sqrt{x}$
and
${g}^{-} 1 \left(x\right) : \left[0 , \infty\right) \to \left(- \infty , 0\right] , {g}^{-} 1 \left(x\right) = - \sqrt[4]{x}$

Note that in each of the above cases, we find that the needed equalities of $f \left({f}^{- 1} \left(x\right)\right) = {f}^{- 1} \left(f \left(x\right)\right) = x$ and $g \left({g}^{-} 1 \left(x\right)\right) = {g}^{-} 1 \left(g \left(x\right)\right) = x$ are satisfied without ambiguity due to the restricted domains.

Graphically, we can also see why this is necessary. If $\left({x}_{0} , {y}_{0}\right)$ is a point on the graph of $y = f \left(x\right)$, then $\left({y}_{0} , {x}_{0}\right)$ should be a point on the graph of $y = {f}^{-} 1 \left(x\right)$. We can find the graph of $y = {f}^{-} 1 \left(x\right)$ by reflecting the graph $y = f \left(x\right)$ about the line $y = x$. If $y = f \left(x\right)$ is not $1 - 1$, however, it will not pass the horizontal line test, meaning ${f}^{-} 1 \left(x\right)$ will not pass the vertical line test, and thus will not be a function.

This is also why when solving ${x}^{2} = a$, we get $x = \pm \sqrt{a}$, as there is no non-multivalued function which is the inverse of ${x}^{2}$ on $\mathbb{R}$.

While we do not arrive there by multiplying by differentials in the typical manner for reciprocals, we can show that $\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$ via the chain rule .

Suppose $y = f \left(x\right)$. Differentiating with respect to $y$, we get

$\frac{d}{\mathrm{dy}} y = \frac{d}{\mathrm{dy}} f \left(x\right)$

$\implies 1 = \frac{\mathrm{dx}}{\mathrm{dy}} \frac{\mathrm{df} \left(x\right)}{\mathrm{dx}}$ (chain rule)

$\implies 1 = \frac{\mathrm{dx}}{\mathrm{dy}} \frac{\mathrm{dy}}{\mathrm{dx}}$

$\therefore \frac{\mathrm{dx}}{\mathrm{dy}} = \frac{1}{\frac{\mathrm{dy}}{\mathrm{dx}}}$

Oct 13, 2016

Disambiguation:

We are faithful to standard notations that involve $. . {\left(\right)}^{- 1}$ ,

${x}^{- 1} = \frac{1}{x} .$ Here, the operand is a prefix. This a particular case of

${x}^{- n} = \frac{1}{x} ^ n$.

With prefix as a function operator and suffix as operand, we have

${\sin}^{- 1} x$ restricted to the principal value of

the angle $\in \left[- \frac{\pi}{2} , \frac{\pi}{2}\right]$, whose sine is x.

Here sin is a prefix function operator and x is the suffix operand.

If $y = f \left(x\right)$,

$x = {f}^{- 1} \left(y\right)$

Here again, f is function prefix and y is the operand.

Commutative law is not applicable to the chain operator $f {f}^{- 1}$.

$f {f}^{- 1}$ is not equivalent to ${f}^{- 1} f$.

A MON AVIS: If y = f(x) and y is locally bijective ( in epsilon-

neighborhood ), the inverse relation is

$x = {f}^{- 1} \left(y\right)$.

For example,

if $y = \sin x \in \left[- 1 , 1\right]$, I define x piecewise as

${f}^{- 1} \left(y\right) = x = k \pi + \left(- 1\right) {\sin}^{- 1} y$, for

$k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots ,$,

with y cyclically $\in \left[- 1 , 1\right]$.

Now, x $\rightarrow \leftarrow y$, making y a locally bijective function.

In our problem here,

for $y = {x}^{2} \in \left[0 , \infty\right) , x \in \left(- \infty , \infty\right)$,

the inverse is defined piecewise as follows.

${f}^{- 1} \left(y\right) = x$

$= - \sqrt{y} , x \in \left(- \infty , 0\right]$

$= \sqrt{y} , x \in \left[0 , \infty\right)$.

Likewise,

for $y = {x}^{4} , \in \left[0 , \infty\right) , x \in \left(- \infty , \infty\right)$,

the inverse is defined as follows.

${f}^{- 1} \left(y\right) = x$

$= - \sqrt{\sqrt{y}} , x \in \left(- \infty , 0\right]$

$= \sqrt{\sqrt{y}} , x \in \left[0 , \infty\right)$.

In either case, this piecewise x is differentiable everywhere for

x'=$\frac{\mathrm{dx}}{\mathrm{dy}}$ and

it can also be verified that $x ' = \frac{1}{y '}$.

It is important, that

the graphs of $y = f \left(x\right) \mathmr{and} x = {f}^{- 1} \left(y\right)$ are one and the same.

So,, it is easy to see that the horizontal test and

vertical test are irrelevant.