Finding the range of values of x where a curve has a negative gradient

AI Thread Summary
The discussion focuses on finding the range of values for x where the curve defined by the equation dy/dx = 3px^2 - m has a negative gradient. The correct approach involves solving the inequality 3px^2 - m < 0, leading to the conclusion that x must be within the range -√(m/3p) < x < √(m/3p). Participants confirm that both p and m are positive, allowing for the determination of this range. The importance of understanding inequalities and their graphical representations is emphasized, suggesting that additional help may be beneficial for those struggling with the concepts. Ultimately, the correct range for x is established as -√(m/3p) < x < √(m/3p).
Natasha1
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Homework Statement
Curve C has equation y = px^3 - mx where p and m are positive integers.
Relevant Equations
Find the range of values of x, in terms of p and m, for which the gradient of C is negative.
dy/dx = 3px^2 - m

Where do I go from here please?
 
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Is this the right approach?

3px^2 - m < 0

3px^2 < m

x^2 < m/3p

x < Square root of m/3p

Is this correct please?
 
Natasha1 said:
Is this the right approach?

3px^2 - m < 0

What do I do here?
This is the right approach. ##p## and ##m## are both positive, so you can work out where the slope ##dy\over dx## is negative.
 
BvU said:
This is the right approach. ##p## and ##m## are both positive, so you can work out where the slope ##dy\over dx## is negative.
Am I correct then when I write this?

3px^2 - m < 0

3px^2 < m

x^2 < m/3p

x < Square root of m/3p
 
Make a plot for e.g. ##m = 3## and ##p=1## to convince yourself ...
 
Natasha1 said:
x^2 < m/3p

x < Square root of m/3p
And check this last step :mad: !
 
Are there two solutions plus (Square root of m/3p) and minus (Square root of m/3p)
 
Natasha1 said:
Are there two solutions plus (Square root of m/3p) and minus (Square root of m/3p)
You have a range for ##x##. Let's write ##a = m/3p##, so that ##a## is some positive number. We have the equation $$x^2 < a$$ What does that say about ##x##? Try drawing a graph of the function ##x^2##, with the line ##y = a## marked.
 
It has two solutions
 
  • #10
Natasha1 said:
It has two solutions
No. You have an inequality. Inequalities typically have a range of solutions.
 
  • #11
What's the solution then, I am stuck.
 
  • #12
Natasha1 said:
What's the solution then, I am stuck.
Let's assume ##a = 1##. What does ##x^2 < 1## tell you?
 
  • #13
That x < + sqrt 1 or x < - sqrt 1
 
  • #14
Natasha1 said:
That x < + sqrt 1 or x < - sqrt 1
Is that what you see on your graph?

E.g. For ##x = -2##, we have ##x < -1## yet ##x^2 = 4 > 1##.
 
  • #15
Never mind, I give up.
 
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  • #16
I need to see the answer to understand where I can't go
 
  • #17
Natasha1 said:
Never mind, I give up.
I recognise your problems with these concepts. You may want to consider a private tutor if grasping the basics of mathematics is important to you. We may not be able to do enough on a forum like this. I can't stand over you and help you draw a graph, for example.
 
  • #18
Natasha1 said:
I need to see the answer to understand where I can't go
Let me give you a basic result of mathematics: $$x^2 < 1$$ is equivalent to $$-1 < x < 1$$
If that's a struggle, then perhaps you need help from someone who has training and knowldege in maths education at this level.
 
  • #19
Where are I going from from here

x^2 < m/3p

x < Square root of m/3p
 
  • #20
Is the answer

- Square root of m/3p < x < + Square root of m/3p
 
  • #21
Natasha1 said:
Is the answer

- Square root of m/3p < x < + Square root of m/3p
Yes.
 
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  • #22
hallelujah
 
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