The function f(x) = x^3 - 3x is.... | vedclass

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(A) Given the function $f(x) = x^3 - 3x$. First,we find the derivative of the function with respect to $x$: $f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3

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Increasing in $(-\infty, -1) \cup (1, \infty)$ and decreasing in $(-1, 1)$

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(A) Given the function $f(x) = x^3 - 3x$. First,we find the derivative of the function with respect to $x$: $f'(x) = \frac{d}{dx}(x^3 - 3x) = 3x^2 - 3$. Factoring the derivative,we get: $f'(x) = 3(x^2 - 1) = 3(x - 1)(x + 1)$. To determine the intervals of increase and decrease,we find the critical points by setting $f'(x) = 0$: $3(x - 1)(x + 1) = 0 \implies x = 1, x = -1$. These points divide the real number line into three intervals: $(-\infty, -1)$,$(-1, 1)$,and $(1, \infty)$. $1$. For $x \in (-\infty, -1)$,choose $x = -2$: $f'(-2) = 3((-2)^2 - 1) = 3(4 - 1) = 9 > 0$. Thus,$f(x)$ is increasing. $2$. For $x \in (-1, 1)$,choose $x = 0$: $f'(0) = 3(0^2 - 1) = -3 $3$. For $x \in (1, \infty)$,choose $x = 2$: $f'(2) = 3(2^2 - 1) = 3(3) = 9 > 0$. Thus,$f(x)$ is increasing. Therefore,$f(x)$ is increasing in $(-\infty, -1) \cup (1, \infty)$ and decreasing in $(-1, 1)$.

The function $f(x) = x^3 - 3x$ is....

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