- #1
Peter G.
- 442
- 0
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!
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!