Given piecewise function f [x]:
f[x_] = Piecewise[{{1/x, x < 0}, {x^2, x >= 0}}]
Another function g [x]:
g[x_] = x + 1
Want to calculate g[f[x]]
get the result is:
How can I obtain the correct answer as shown in the following figure:
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
Use
g[f[x]] // PiecewiseExpand
How to make the definition field display as: x>=0?
I do not know offhand without writing special code to post-process it. But these are really the same. i.e. True in this case is the same as x>=0, because True means for all other possible values of x, which is x>=0.
So Mathematically they are the same. If you want x>=0 to show there instead of True just for display purposes, I think special code might be needed to do this. I could not see an option in PiecewiseExpand to change this when I just looked. May be someone will have an idea how to do this easily.
-
$\begingroup$ The definition field corresponding to 1+x ^ 2 in the obtained results is true. How to make the definition field display as: x>=0? $\endgroup$– csn899Commented Jun 5, 2023 at 23:38
$\begingroup$
$\endgroup$
Clear["Global`*"]
f[x_] = Piecewise[{{1/x, x < 0}, {x^2, x >= 0}}];
g[x_] = x + 1;
As shown by Nasser,
h[x_] = g[f[x]] // PiecewiseExpand
Post-processing,
h2[x_] = ReplacePart[
h[x], {1 -> Append[h[x][[1]], {h[x][[2]], Not[h[x][[1, 1, -1]]]}],
2 -> 0}]
answered Jun 6, 2023 at 2:44