The equation x^3+x-1=0 has | vedclass

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(C) Let $f(x) = x^3 + x - 1$. Since $f(0) = -1  0$,by the Intermediate Value Theorem,there exists at least one real root $c$ in th

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exactly one real root.

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(C) Let $f(x) = x^3 + x - 1$. Since $f(0) = -1  0$,by the Intermediate Value Theorem,there exists at least one real root $c$ in the interval $(0, 1)$. Alternatively,consider the derivative $f'(x) = 3x^2 + 1$. Since $3x^2 + 1 > 0$ for all $x \in \mathbb{R}$,$f(x)$ is a strictly increasing function. $A$ strictly increasing cubic function crosses the $X$-axis exactly once. Therefore,the equation has exactly one real root.

The equation $x^3+x-1=0$ has

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