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

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(A) The force is given by $F = 2x^2 - 3x - 2$. Equilibrium occurs when $F = 0$.$2x^2 - 3x - 2 = 0$$2x^2 - 4x + x - 2 = 0$$2x(x - 2) + 1(x - 2) = 0$$(2

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

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(A) The force is given by $F = 2x^2 - 3x - 2$. Equilibrium occurs when $F = 0$.$2x^2 - 3x - 2 = 0$$2x^2 - 4x + x - 2 = 0$$2x(x - 2) + 1(x - 2) = 0$$(2x + 1)(x - 2) = 0$Thus,equilibrium positions are $x = -1/2$ and $x = 2$.To determine stability,we check the derivative of force: $dF/dx = 4x - 3$.For stable equilibrium,$dF/dx At $x = -1/2$: $dF/dx = 4(-1/2) - 3 = -2 - 3 = -5 At $x = 2$: $dF/dx = 4(2) - 3 = 8 - 3 = 5 > 0$. This is unstable equilibrium.Therefore,$x = -1/2$ is a position of stable equilibrium.

If $F = 2x^2 - 3x - 2$,then choose the correct option.

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