Answer
$x_1=-0.424ft$ unstable equilibrium
$x_2=0.590ft$ stable equilibrium
Work Step by Step
The required equilibrium positions can be determined as follows:
Given that, $V=4x^3-x^2-3x+10$
At equilibrium, $\frac{dV}{dx}=0$
$\implies \frac{d(4x^3-x^2-3x+10)}{dx}=12x^2-2x-3=0$ eq(1)
This simplifies to:
$x_1=-0.424$
and $x_2=0.590$
Now we take second derivative of eq(1)
For $x_1=-0.424$ $\frac{d^2V}{dx^2}=24(-0.424)-2=-12.176\lt 0$
which means unstable equilibrium
For $x_2=0.590$ $\frac{d^2V}{dx^2}=24(0.590)-2=12.16\gt 0$
which means stable equilibrium.
