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

  • Thread starter Natasha1
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  • #1
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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?
 

Answers and Replies

  • #2
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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?
 
  • #3
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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.
 
  • #4
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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
 
  • #5
BvU
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Make a plot for e.g. ##m = 3## and ##p=1## to convince yourself ...
 
  • #6
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x^2 < m/3p

x < Square root of m/3p
And check this last step :mad: !
 
  • #7
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Are there two solutions plus (Square root of m/3p) and minus (Square root of m/3p)
 
  • #8
PeroK
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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.
 
  • #10
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It has two solutions
No. You have an inequality. Inequalities typically have a range of solutions.
 
  • #11
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What's the solution then, I am stuck.
 
  • #12
PeroK
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What's the solution then, I am stuck.
Let's assume ##a = 1##. What does ##x^2 < 1## tell you?
 
  • #13
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That x < + sqrt 1 or x < - sqrt 1
 
  • #14
PeroK
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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
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Never mind, I give up.
 
  • #16
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I need to see the answer to understand where I can't go
 
  • #17
PeroK
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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
PeroK
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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
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Where are I going from from here

x^2 < m/3p

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

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

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

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