Value that gives more than one stationary point

[maxim07]

Homework Statement:

What value of z makes the curve have more than one stationary point

Relevant Equations:

f(x) = 2x^3 -12x^2 + zx + 30

Here is my attempt at a solution

I don’t know where to go from here, as the equation is a quadratic it will have two solutions which means there will be two stationary points, but I don’t know how to solve for z.

 

[PeroK]

I don’t know where to go from here, as the equation is a quadratic it will have two solutions

Does a quadratic always have two solutions?

 

[maxim07]

Not if both brackets are the same

 

[PeroK]

Not if both brackets are the same

I have no idea what you mean by that. What brackets?

 

[FactChecker]

I don’t know where to go from here, as the equation is a quadratic it will have two solutions which means there will be two stationary points

That is only guaranteed if you allow complex solutions. When the problem expects only real solutions, the quadratic formula must have non-negative numbers inside the square root. That should allow you to get the restrictions on z.

 

[maxim07]

Not if both brackets are the same

I mean sometimes if you factor a quadratic the two bracketed terms you end up with are the same so you only have one solution

 

[PeroK]

I mean sometimes if you factor a quadratic the two bracketed terms you end up with are the same so you only have one solution

Can you always factorise a quadratic? What about $x^2 + 1$?

 

[FactChecker]

I mean sometimes if you factor a quadratic the two bracketed terms you end up with are the same so you only have one solution

You need to use the quadratic formula. That will give you the factors, whether they are real, complex, integer, irrational, etc.

 

[maxim07]

using the fact that the term under the square root can not be negative, this is what I get

 

[Mark44]

So to summarize, there will be two stationary points if z < 24.

What happens if z = 24? Or if z > 24?

Comment - the usual way of writing the derivative of f at a value x is f'(x), not f(x)'. IOW, the "prime" symbol (an apostrophe) goes immediately after the function name.

 

[maxim07]

If z = 24 you get sqrt(0) , if z>24 you get a negative in the square root, neither of which are possible

 

[Mark44]

If z = 24 you get sqrt(0) , if z>24 you get a negative in the square root, neither of which are possible

No, both these cases are possible. If z = 24, there is only one stationary point. If z > 24, there are no stationary points. In the latter case, this means that the graph of $f(x) = 2x^3 - 12x^2 + zx + 30$ is always increasing; i.e., has no points at which the tangent line is horizontal.

 

[maxim07]

So if z =24 we get √0 so there is only one x solution and so one stationary point, if z > 24 we get √- so there are no solutions and no stationary points

 

[FactChecker]

So if z =24 we get √0 so there is only one x solution and so one stationary point, if z > 24 we get √- so there are no solutions and no stationary points

Yes, on the real line. When you start to talk about complex numbers, there are complex solutions for the square root of negative numbers, but that is another subject.