If F = 2x^2 - 3x - 2,then select the correct | vedclass

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(C) For equilibrium,the force $F$ must be zero.$F = 2x^2 - 3x - 2 = 0$Factoring the quadratic equation: $2x^2 - 4x + x - 2 = 0 \Rightarrow 2x(x - 2) +

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$x = -1/2$ is the position of stable equilibrium.

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(C) For equilibrium,the force $F$ must be zero.$F = 2x^2 - 3x - 2 = 0$Factoring the quadratic equation: $2x^2 - 4x + x - 2 = 0 \Rightarrow 2x(x - 2) + 1(x - 2) = 0 \Rightarrow (2x + 1)(x - 2) = 0$.Thus,the equilibrium positions are $x = 2$ and $x = -1/2$.The potential energy $U$ is related to force by $F = -dU/dx$,so $dU/dx = -F$.The condition for stable equilibrium is that the potential energy $U$ is minimum,which implies $d^2U/dx^2 > 0$.Since $dU/dx = -F$,we have $d^2U/dx^2 = -dF/dx$.Calculating the derivative: $dF/dx = d/dx(2x^2 - 3x - 2) = 4x - 3$.Therefore,$d^2U/dx^2 = -(4x - 3) = 3 - 4x$.At $x = -1/2$: $d^2U/dx^2 = 3 - 4(-1/2) = 3 + 2 = 5 > 0$. Since the second derivative is positive,$x = -1/2$ is a position of stable equilibrium.At $x = 2$: $d^2U/dx^2 = 3 - 4(2) = 3 - 8 = -5 < 0$. Since the second derivative is negative,$x = 2$ is a position of unstable equilibrium.

If $F = 2x^2 - 3x - 2$,then select the correct statement.

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