2{x^3} - 6x + 5 is an increasing function if | vedclass

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(C) Let $f(x) = 2{x^3} - 6x + 5$. To find the intervals where the function is increasing,we calculate the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2{

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(C) Let $f(x) = 2{x^3} - 6x + 5$. To find the intervals where the function is increasing,we calculate the derivative $f'(x)$. $f'(x) = \frac{d}{dx}(2{x^3} - 6x + 5) = 6{x^2} - 6$. $A$ function is increasing when $f'(x) > 0$. So,$6{x^2} - 6 > 0$. Dividing by $6$,we get ${x^2} - 1 > 0$. This can be factored as $(x - 1)(x + 1) > 0$. Using the sign scheme method,the inequality holds when $x > 1$ or $x Thus,the function is increasing for $x \in (-\infty, -1) \cup (1, \infty)$.

$2{x^3} - 6x + 5$ is an increasing function if

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