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

2 Answers
Oct 11, 2016

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.


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)


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.


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)

Oct 13, 2016

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.