The set of values of x for which f(x)=3x^4-8x | vedclass

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(B) To find the intervals where $f(x)$ is increasing,we find the derivative $f'(x)$ and set $f'(x) > 0$. Given $f(x) = 3x^4 - 8x^3 - 6x^2 + 24x - 12$.

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$(-1, 1) \cup (2, \infty)$

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(B) To find the intervals where $f(x)$ is increasing,we find the derivative $f'(x)$ and set $f'(x) > 0$. Given $f(x) = 3x^4 - 8x^3 - 6x^2 + 24x - 12$. $f'(x) = 12x^3 - 24x^2 - 12x + 24$. Factor out $12$: $f'(x) = 12(x^3 - 2x^2 - x + 2)$. Factor by grouping: $f'(x) = 12[x^2(x - 2) - 1(x - 2)] = 12(x^2 - 1)(x - 2) = 12(x - 1)(x + 1)(x - 2)$. For $f(x)$ to be increasing,$f'(x) > 0$,so $(x - 1)(x + 1)(x - 2) > 0$. Using the wavy curve method (sign scheme) for the critical points $x = -1, 1, 2$: For $x \in (-1, 1)$,$f'(x) > 0$. For $x \in (2, \infty)$,$f'(x) > 0$. Thus,the function is increasing on $(-1, 1) \cup (2, \infty)$.

The set of values of $x$ for which $f(x)=3x^4-8x^3-6x^2+24x-12$ is an increasing function,is

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