       Re: finding inverses of functions

• To: mathgroup at smc.vnet.net
• Subject: [mg126441] Re: finding inverses of functions
• From: "djmpark" <djmpark at comcast.net>
• Date: Thu, 10 May 2012 05:00:52 -0400 (EDT)
• Delivered-to: l-mathgroup@mail-archive0.wolfram.com
• References: <4767427.124602.1336550815297.JavaMail.root@m06>

```An interesting problem. I'm going to take your "e" to be Exp. Let's define
the y function:

Clear[x, y];
y[x_] := 3 x^3 + 2 E^(2 x);

The function looks to be monotonically increasing.

Plot[y[x], {x, -2, 2}]

And we can verify it.

ForAll[x, D[y[x], x] > 0]
Resolve[%, Reals]
...
True

We will solve a differential equation and need an initial condition:

y
2

Write the inverse slope in terms of x[y] and solve the differential
equation.

Clear[x]
x'[y] == 1/(D[y[x], x] /. x -> x[y]);
xsol = DSolve[%, x, y][[1, 1]]

x -> Function[{y}, InverseFunction[2 E^(2 #1) + 3 #1^3 &][y + C]]

Solve the initial condition for C.

x == 0 /. xsol;
Solve[%, C][[1, 1]]
C -> 0

We can now define the x function. It is in terms of an InverseFunction but
easily evaluable.

x[y_] = x[y] /. (xsol /. C -> 0)
InverseFunction[2 E^(2 #1) + 3 #1^3 &][y]

We can check numerically that the function is inverse, at least for small or
exact values of x.

x[y[x]] == x;
% /. x -> 10
True

(I wish I knew a method to show this symbolically.)

We can plot the y function, inverse x function and their composition to
check, at least numerically, the proper relation.

Show[
{Plot[y[x], {x, -2, 2}, PlotStyle -> Black],
Plot[x[y], {y, -20, 20}, PlotStyle -> Blue],
Plot[y[x[z]], {z, -10, 10}, PlotStyle -> Red]},
AspectRatio -> Automatic,
PlotRange -> 10,
Axes -> False,
Frame -> True]

David Park
djmpark at comcast.net
http://home.comcast.net/~djmpark/index.html

From: John Accardi [mailto:accardi at accardi.com]

Hello,  I am trying to get Mathematica to find the inverse of:

y = 3x^3 + 2e^(2x)  (which I know is invertible)

InverseFunction only seems to give inverses of built-ins, like Sine.

I tried:

Solve[ y == 3x^3 + 2e^(2x), x ] but get a message that Solve does not have
methods suitable.  (Solve works for simpler functions, however.)

Any ideas?  Thanks.

```

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