The second derivative. If f" is <0, it is unstable
a) f=4x-2x^3
f'=4-6x^2
f"=-12x unstable at zero,
now where is f=0?
0=x(4-2x^2) x=0 unstable
x=1.41 unstable
x=-1.41 stable
b. F = x2 + x − 6
f'=2x+1
f"=2 always stable.
https://en.wikipedia.org/wiki/Mechanical_equilibrium
wondering where you got these "answers", as they are wrong.
zeroes: 0=(x-3)(x+2)
x=3 stable, x=-2 stable
c. F = 5x − 9.
f'=5
f"=0 neutral
zeroes: x=1.8
For each of the following functions describing the net force on a particle, determine any and all positions of equilibrium and characterize each as stable or unstable. (a) F = 4x − 2x3, (b) F = x2 + x − 6, (c) F = 5x − 9.
Answer: a. x = 0, unstable; x = ±1.41, both stable
b. x = −3, stable; x = 2, unstable
c. x = −1.8, unstable
Please help me figure out how to get the answer. Thank you!
2 answers
The teacher gives us the answer, but still thank you!