Maths Help!

[LoveToStack]

I'm not sure if my method is correct here, just wondering if anyone can confirm that this is the right way to do things.
Basically I'm trying to prove that:

y=2x^3 +3x^2 +4x -5

has no stationary points. To find stationary points I've differentiated to get:

dy/dx = 6x^2 +6x +4

Then I've made the statement that stationary points occur where dy/dx= 0 and got:

6x^2 +6x +4=0
3x^2 +3x +2=0

Then that doesn't factorise so I've used the discriminant, b^2 -4ac, to determine whether 3x^2 +3x +2 has any real roots (this is where I'm a bit sketchy because obviously you wouldn't normally use the discriminant here but I'm doing so because dy/dx doesn't factorise). Using that, I determined that b^2 -4ac = -15 concluding that there are no real roots (because its <0) so basically it doesn't factorise. So based on the fact that the derivative does not cross the x-axis at any point, I'm concluding that the main curve has no stationary points - is this a valid statement to make? All help appreciated!

[Bun]

I'm not sure if my method is correct here, just wondering if anyone can confirm that this is the right way to do things.
Basically I'm trying to prove that:

y=2x^3 +3x^2 +4x -5

has no stationary points. To find stationary points I've differentiated to get:

dy/dx = 6x^2 +6x +4

Then I've made the statement that stationary points occur where dy/dx= 0 and got:

6x^2 +6x +4=0
3x^2 +3x +2=0

Then that doesn't factorise so I've used the discriminant, b^2 -4ac, to determine whether 3x^2 +3x +2 has any real roots (this is where I'm a bit sketchy because obviously you wouldn't normally use the discriminant here but I'm doing so because dy/dx doesn't factorise). Using that, I determined that b^2 -4ac = -15 concluding that there are no real roots (because its <0) so basically it doesn't factorise. So based on the fact that the derivative does not cross the x-axis at any point, I'm concluding that the main curve has no stationary points - is this a valid statement to make? All help appreciated!

if i remember correctly, if you can't factorise dy/dx then it has no real roots, i think.

Edited by Nicola (Forum Moderator): Posts merged due to forum lag.

[CHA!NGANG]

Does a stationary point have to be a root? A root is where it crosses the x-axis and a stationary point is a min, max or point of inflection so it doesn't need to cross the x-axis anyway. Not sure though could be talking rubbish.

[Soka]

Does a stationary point have to be a root? A root is where it crosses the x-axis and a stationary point is a min, max or point of inflection so it doesn't need to cross the x-axis anyway. Not sure though could be talking rubbish.

Your statement is correct, but he said a root of the derivative.. so i think you've misread his statement.

Lovetostack, your answer seems fine and correct to me.

[Bun]

yeah er listen to soka, he does maths at oxford i thinkkkkkkkkkkkkk?

[CHA!NGANG]

Your statement is correct, but he said a root of the derivative.. so i think you've misread his statement.

Lovetostack, your answer seems fine and correct to me.

Ah yeah of course. I was thinking of the origional equation for somereason not the derivative. Yeah I think that he's right then.

[Axel]

Well whatever x is, it's between 0 and -1.

[Soka]

Well whatever x is, it's between 0 and -1.

Do you mean the solution to 3x^2 + 3x + 2? Because it will not be a real number, it will be a pair of complex conjugates. But thats irrelevant anyway to the original question as Lovetostack simply needs to show the discriminant is less than zero, hence no real roots to the solution thus no stationary points.

Having read through the statement again lovetostack, you don't need to include:

So based on the fact that the derivative does not cross the x-axis at any point

because that is not the reason, as the derivative is the gradient of the line, at a given point of x, not a point to cross the x-axis. you've already stated the correct reason (b^2-4ac <0 etc.). If you want me to explain why this is the case, i'll try to explain so it helps you.

[DieselShaq]

Looks like I came a bit late and Soka already solved this. But yes, he is correct and so is your method.

[LoveToStack]

Thanks very much everyone, especially Soka +rep.