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

[Natasha1]

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?

 

[Natasha1]

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?

 

[BvU]

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.

 

[Natasha1]

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

 

[BvU]

Make a plot for e.g. $m = 3$ and $p=1$ to convince yourself ...

 

[BvU]

x^2 < m/3p

x < Square root of m/3p

And check this last step !

 

[Natasha1]

Are there two solutions plus (Square root of m/3p) and minus (Square root of m/3p)

 

[PeroK]

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.

 

[Natasha1]

It has two solutions

 

[PeroK]

It has two solutions

No. You have an inequality. Inequalities typically have a range of solutions.

 

[Natasha1]

What's the solution then, I am stuck.

 

[PeroK]

What's the solution then, I am stuck.

Let's assume $a = 1$. What does $x^2 < 1$ tell you?

 

[Natasha1]

That x < + sqrt 1 or x < - sqrt 1

 

[PeroK]

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 $x^2 = 4 > 1$.

 

[Natasha1]

Never mind, I give up.

 

[Natasha1]

I need to see the answer to understand where I can't go

 

[PeroK]

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.

 

[PeroK]

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.

 

[Natasha1]

Where are I going from from here

x^2 < m/3p

x < Square root of m/3p

 

[Natasha1]

Is the answer

- Square root of m/3p < x < + Square root of m/3p

 

[PeroK]

Is the answer

- Square root of m/3p < x < + Square root of m/3p

Yes.

 

[Natasha1]

hallelujah