1
$\begingroup$
Clear["Global`*"]
f[x_] = Log[10, x + a/x - 2]
FunctionDomain[{f[x], a > 0}, x] // Reduce[#, a] &

USE

In[22]:= Clear["Global`*"]
f[x_] = Log[10, x + a/x - 2]
FunctionDomain[{f[x], a > 0}, x] // Reduce[#, a] &

Out[23]= Log[-2 + a/x + x]/Log[10]

Out[24]= (0 < x <= 2 && a > 2 x - x^2) || (x > 2 && a > 0)

The answer above is incorrect

the correct answer is as follows:

(0 < a <= 1 && (0 < x < 1 - Sqrt[1 - a] || x > 1 + Sqrt[1 - a])) || (a > 1 && x > 0)

$\endgroup$

1 Answer 1

4
$\begingroup$
FunctionDomain[{Log[10, x + a/x - 2], a > 0}, x]// Reduce[#, x] &
(*(0 < a <=1 && (0 < x < 1 - Sqrt[1 - a] || x > 1 + Sqrt[1 - a])) || (a > 1&&x > 0)*)

gives the "correct asnwer".

By the way RegionPlot[...] shows no difference between the two conditions // Reduce[#, a] & and // Reduce[#, x] & !

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1
  • 1
    $\begingroup$ if the two domains are given by {fd1, fd2} then {rgn1, rgn2} = ImplicitRegion[#, {a, x}] & /@ {fd1, fd2}; and RegionEqual[rgn1, rgn2] evaluates to True $\endgroup$
    – Bob Hanlon
    Jun 14, 2023 at 16:52

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