The equation $x^{3} - 3 x + 4 = 0$ has only one real root. What is its first approximate value as obtained by the method of false position in (-3, -2)?

-2.125

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2.125

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-2.812

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2.812

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Solution

The correct option is A -2.125
$x_{n} = x_{1} - \frac{x_{n - 1} - x_{1}}{f (x_{n - 1}) - f (x_{1})} f (x_{1})$
Here $x_{n - 1} = - 2$ and $x_{1} = - 3$
$f (x_{1}) = f (- 3) f (- 3) = {(- 3)}^{3} - 3 (- 3) + 4 = - 14 f (x_{n - 1}) = f (- 2) f (- 2) = {(- 2)}^{3} - 3 (- 2) + 4 = 2 x_{n} = - 3 - \frac{- 2 + 3}{2 + 14} (- 14) x_{n} = - 2.125$
So option $A$ is correct

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