Question

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

Answer 1

`dx/dy = 1/(dy/dx)`

Explanation:

Assuming we are looking for the inverse functions of `f(x) = x^2` and `g(x) = x^4`, i.e., `f^(-1)(x)` and `g^(-1)(x)` such that

`f(f^(-1)(x))=f^(-1)(f(x)) = x`
and
`g(g^-1(x)) = g^-1(g(x)) = 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 `RR`, should we have `f^-1(4)=2` or `f^(-1)(4)=-2`?

Again, assuming we are only considering real-valued functions, we can partition `RR` 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 RR^+` such that `y^2=x`. Similarly, `root(4)(x)` will represent the unique `y in RR^+` such that `y^4 = x`.

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

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If we have
`f(x):[0, oo)->[0, oo), f(x) = x^2`
and
`g(x):[0, oo)->[0, oo), g(x) = x^4`

then we can find their inverse functions as

`f^-1(x):[0, oo)->[0, oo), f^-1(x) = sqrt(x)`
and
`g^-1(x):[0, oo)->[0, oo), g^-1(x) = root(4)(x)`

---

If we have
`f(x):(-oo,0]->[0, oo), f(x) = x^2`
and
`g(x):(-oo,0]->[0, oo), g(x) = x^4`

then we can find their inverse functions as

`f^-1(x):[0, oo)->(-oo,0], f^-1(x) = -sqrt(x)`
and
`g^-1(x):[0, oo)->(-oo,0], g^-1(x) = -root(4)(x)`

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

Graphically, we can also see why this is necessary. If `(x_0, y_0)` is a point on the graph of `y=f(x)`, then `(y_0, x_0)` should be a point on the graph of `y = f^-1(x)`. We can find the graph of `y=f^-1(x)` by reflecting the graph `y=f(x)` about the line `y=x`. If `y=f(x)` is not `1-1`, however, it will not pass the horizontal line test, meaning `f^-1(x)` 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 = +-sqrt(a)`, as there is no non-multivalued function which is the inverse of `x^2` on `RR`.

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While we do not arrive there by multiplying by differentials in the typical manner for reciprocals, we can show that `dx/dy = 1/(dy/dx)` via the chain rule .

Suppose `y = f(x)`. Differentiating with respect to `y`, we get

`d/dyy = d/dyf(x)`

`=> 1 = dx/dy (df(x))/dx` (chain rule)

`=> 1 = dx/dy dy/dx`

`:. dx/dy = 1/(dy/dx)`

Answer 2

Disambiguation:

We are faithful to standard notations that involve `..()^(-1)` ,

`x^(-1)=1/x.` Here, the operand is a prefix. This a particular case of

`x^(-n)=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 [-pi/2, pi/2]`, whose sine is x.

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

If `y = f(x)`,

`x = f^(-1)(y)`

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

Commutative law is not applicable to the chain operator `f f^(-1)`.

`ff^(-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)(y)`.

For example,

if `y = sin x in [-1, 1]`, I define x piecewise as

`f^(-1)(y) = x = kpi+(-1)sin^(-1)y`, for

`k = 0, +-1, +-2, +-3, ...,`,

with y cyclically `in [-1, 1]`.

Now, x `rarrlarry`, making y a locally bijective function.

In our problem here,

for `y = x^2 in [0, oo), x in ( -oo, oo )`,

the inverse is defined piecewise as follows.

`f^(-1)(y)=x`

`= -sqrt y, x in (-oo, 0]`

`= sqrt y, x in [0, oo)`.

Likewise,

for `y = x^4, in [0, oo), x in ( -oo, oo )`,

the inverse is defined as follows.

`f^(-1)(y)=x`

`= -sqrt sqrt y, x in (-oo, 0]`

`= sqrt sqrt y, x in [0, oo)`.

In either case, this piecewise x is differentiable everywhere for

x'=`(dx)/(dy)` and

it can also be verified that `x'=1/(y')`.

It is important, that

the graphs of `y = f(x) and x = f^(-1)(y)` are one and the same.

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

vertical test are irrelevant.