[?] Solve the following inequality with W|A?

Questions
,
1

Consider the following W|A input:

(x^3+2x^2+3x)^(1/3)-x-1<=0

Solution Wolfram → x>=0 ??? why?

Solution is : -inf <= x <= +inf

x = -2 is ok!!! -2 <=0

2

Hello Faustino,

I don’t see that x = -2 is OK. For x = -2 the term from which you have to take root 1/3 is negative. The number you get after taking the root will be complex.

3

Just to illustrate:

In[1]:= t = (x^3 + 2 x^2 + 3 x)^(1/3) - x - 1
Out[1]= -1 - x + (3 x + 2 x^2 + x^3)^(1/3)
    
In[2]:= N[t /. x -> -2]
Out[2]= 1.90856 + 1.57367 I
4

No, with x=-2 → -0,817<=0

5

Looks like you are assuming that 1 + (-6)^(1/3) is the same as 1 - 6^(1/3). It is not.

6

To indicate the issue noted by @RohitNamjoshi, here is the difference in results between using the exponent 1/3 vs using the real-valued CubeRoot function.

Reduce[(x^3 + 2 x^2 + 3 x)^(1/3) - x - 1 <= 0, x]

(* Out[366]= x >= 0 *)

Reduce[CubeRoot[(x^3 + 2 x^2 + 3 x)] - x - 1 <= 0, x]

(* Out[367]= x \[Element] Reals *)
7

You might want to look up Surd (https://reference.wolfram.com/language/ref/Surd.html).

Especially, you might want to look at the “Possible Issues” section where it says:

“On the negative real axis, Surd[x,n] is different from the principal root returned by Power[x,1/n]:”

and look a the following examples. If you look at Surd instead of Power then

1 + Surd[-6,3] == 1 - Surd[6,3]

is actually True. Rather than

1 + (-6)^(1/3) == 1 - 6^(1/3)

which is False. Note that

Reduce[Surd[x^3 + 2 x^2 + 3 x, 3] - x - 1 <= 0, x]

gives the solution that you expect if I am not mistaken.

Cheers,

Marco

8

Sorry Daniel,

your answer only showed up after I submitted my attempt at a reply.

Cheers,
Marco