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

  • Thread starter peter.a
  • Start date
In summary, the conversation discussed using Mathematica to solve equations and express solutions in terms of previously obtained values. The output showed that the solution for k is an inequality, which can be interpreted using mathematical understanding. It was also suggested to think of Mathematica as a calculator with extra buttons rather than a human mathematician.
  • #1
peter.a
21
0
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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  • #2
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
 
  • #3
the solution for x is an inequality, shouldn't the solution for k then also be an inequality ?
 
  • #4
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:
  • #5
You can solve the above equation by hand, do you want to know how?
 

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

1. What does "h'[x]" mean in the equation?

"h'[x]" represents the derivative of the function "h(x)". In other words, it is the rate of change of the function with respect to the variable x.

2. How do you solve for h(x) in terms of k in this equation?

To solve for h(x) in terms of k, we need to first isolate h(x) on one side of the equation. Then, we can take the inverse of any operations that are being applied to h(x) and apply it to both sides of the equation. This will give us the solution of h(x) in terms of k.

3. What is the significance of "k" in this equation?

"k" represents a constant value in this equation. It could represent a specific numerical value or a variable with a specific value. The value of "k" will determine the value of h(x) in the solution.

4. Can this equation have multiple solutions?

Yes, this equation can have multiple solutions for h(x) in terms of k. This means that there can be multiple values of h(x) that satisfy the equation for a given value of k. It is important to check for any restrictions on the domain of the function to ensure all solutions are valid.

5. How can this equation be applied in scientific research?

This equation can be applied in scientific research to find the values of h(x) that satisfy the given conditions. This can be useful in various fields such as physics, chemistry, and biology where functions and their derivatives are used to model real-world phenomena. The solutions can provide insights and understanding into the behavior of these systems.

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