Find Solutions of h'[x] >= 1 in Terms of k

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The discussion focuses on solving the inequality h'[x] >= 1 in terms of k, where h[x] = x^3 + x^2 + k. The user attempts to express the inequalities derived from the function in terms of k but struggles with Mathematica's output, which does not meet their expectations. They note that while Mathematica can provide a solution, it often requires complex manipulation to achieve the desired format. The conversation emphasizes the importance of understanding the underlying mathematics rather than solely relying on software for formatting. Ultimately, the user seeks a simpler approach to using Mathematica for their problem-solving needs.
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h[x_] = x^3 +x^2+ k;
Solve[h[x] == 0, k]
{{k -> -x^2 - x^3}}

Reduce[Abs[h'[x]] >= 1, x]
Re[x] < -(1/2) || (-(1/2) <= Re[x] <= 1/2 && (Im[x] <= -(1/2) Sqrt[1 - 4 Re[x]^2] ||
Im[x] >= 1/2 Sqrt[1 - 4 Re[x]^2])) || Re[x] > 1/2

Now i want to express these inequalities in terms of the value of k i obtained earlier, how can i do this, i have tried the mathematica online resource but i havn't been able to figure it out.

Reduce[Abs[h'[x]] >= 1, k]
gives me the following output:
Abs[-1 + 2 x] >= 1
whereas i want my solution in terms of k
 
Last edited:
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In[1]:= h[x_]:=x^3+x^2+k;Solve[h[x]==0,k]

Out[2]= {{k-> -x^2-x^3}}

In[3]:= Reduce[Abs[h'[x]]≥1&&x∈Reals,x]

Out[3]= x ≤ -1 || x ≥ 1/3

In[4]:= Eliminate[k== -x^2-x^3&&x== -1,x]

Out[4]= k==0

In[5]:= Eliminate[k== -x^2-x^3&&x== 1/3,x]

Out[5]= 27 k == -4
 
the solution for x is an inequality, shouldn't the solution for k then also be an inequality ?
 
I am attempting to show you how to calculate what you need as simply as possible.

I can add on layer after layer after layer of incomprehensible Mathematica-speak so that it will display k<=0 instead of k==0, which is the end point of the inequality,
OR
I can show you much simpler use of Mathematica that you hopefully have some chance of understanding and you then use a little bit of mental mathematical understanding to see how to interpret the result. As a bonus you might be able to see how to use the simpler version for other problems in the near future.

Pick one.

It is even possible that someone, or perhaps even you after dozens or hundreds or thousands of hours of learning the system, can come up with a very simple way of getting exactly what you want for this particular problem. A useful guide for Mathematica is that it takes two to ten times longer to get something that looks pretty close to the way you want it to look than it takes to get the math approximately correct and two to ten times longer than that to get it to look almost exactly the way you want it, or you just give up and use what you have. I suggest spending your time making really sure you have the math correct.

If you think of Mathematica as

a calculator with lots of extra buttons

instead of thinking of it as

a bright mathematician that can tell he is supposed to go in the back room without even being told, figure out what your problem really ought to be instead of what you think it is, figure out the solution to your problem, desktop publish it so it looks like it does in a textbook and then bring it out and show this to you

then I think Mathematica will be more understandable.
 
Last edited:
You can solve the above equation by hand, do you want to know how?
 

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