# Differentiation and Stationary Points

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!

## Answers and Replies

Dick
Science Advisor
Homework Helper
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?

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
Science Advisor
Homework Helper
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.

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 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.

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
Science Advisor
Homework Helper
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?