Differentiation and Stationary Points

[Peter G.]

Hi,

Q: By investigating the stationary points of f(x)= x³+3x²+6x-30 and sketching the curve y=f(x) show that the equation f(x)=0 has only one real solution.

A: Well, I don't understand how I should use both. Plotting the graph, I can clearly spot a solution: x = 1.9319548

I know how to investigate the stationary points. I first found the first derivative: 3x²+6x+6, which had no real solution, so I moved to the second derivative: 6x+6, but I still don't get the connection. How can I use stationary points to define where the cubic will intersect the x-axis.

Any tips?

Thanks!

 

[Dick]

Hi,

Q: By investigating the stationary points of f(x)= x³+3x²+6x-30 and sketching the curve y=f(x) show that the equation f(x)=0 has only one real solution.

A: Well, I don't understand how I should use both. Plotting the graph, I can clearly spot a solution: x = 1.9319548

I know how to investigate the stationary points. I first found the first derivative: 3x²+6x+6, which had no real solution, so I moved to the second derivative: 6x+6, but I still don't get the connection. How can I use stationary points to define where the cubic will intersect the x-axis.

Any tips?

Thanks!

Since there are no real zeros for the derivative then there are no stationary points. Consider the limits as x->infinity and x->-infinity. Can you argue there must be at least one root? Can you argue that there can't be two?

 

[Peter G.]

Ok, I understood: "Since there are no real zeros for the derivative then there are no stationary points." But, I have to find roots in the cubic?

 

[Dick]

Ok, I understood: "Since there are no real zeros for the derivative then there are no stationary points." But, I have to find roots in the cubic?

No, you don't have to find them. You just have to show that there is exactly one of them.

 

[Peter G.]

I've learned about discriminants for quadratics. I did a quick search and there is one for cubics. Another website claimed that + + + -, like in my case, then there is only one real root. Sorry, I've never dealt with cubics

 

[DrewD]

I don't think the question has anything to do with the formula for finding roots to a cubic. The question is asking you to use your knowledge of derivatives and how they affect the shape of a graph. If you look up the "cubic formula", you have not answered the question. Yes, you are correct, but the question was very specific about how you should show the answer.

 

[Peter G.]

Um... ok. So, for example. From the graph I know that the solution is at 1.93. So, if I use the first derivative, I can show that to the left and right, the gradient will always be positive, hence, meaning the line wouldn't ascend/descend, cutting the x-axis again?

 

[Dick]

Um... ok. So, for example. From the graph I know that the solution is at 1.93. So, if I use the first derivative, I can show that to the left and right, the gradient will always be positive, hence, meaning the line wouldn't ascend/descend, cutting the x-axis again?

Yes, you are almost there. But you didn't even need to find the root at 1.93. If x is negative and large then your polynomial is a large negative number. If x is positive and large then your polynomial is a large positive number. There must be at least one root since it has to cross the x-axis someplace, yes?