The function f(x) = 2{x^3} + 18{x^2} - 96x + | vedclass

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(A) To find the intervals where the function $f(x) = 2{x^3} + 18{x^2} - 96x + 45$ is increasing,we calculate its derivative $f'(x)$.$f'(x) = \frac{d}{

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$x \le - 8$ or $x \ge 2$

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(A) To find the intervals where the function $f(x) = 2{x^3} + 18{x^2} - 96x + 45$ is increasing,we calculate its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(2{x^3} + 18{x^2} - 96x + 45) = 6{x^2} + 36x - 96$.For the function to be increasing,we must have $f'(x) \ge 0$.$6{x^2} + 36x - 96 \ge 0$.Dividing the inequality by $6$,we get:${x^2} + 6x - 16 \ge 0$.Factoring the quadratic expression:$(x + 8)(x - 2) \ge 0$.Using the sign scheme method,the expression $(x + 8)(x - 2)$ is greater than or equal to $0$ when $x \le - 8$ or $x \ge 2$.Thus,the function is increasing on the intervals $(-\infty, -8] \cup [2, \infty)$.

The function $f(x) = 2{x^3} + 18{x^2} - 96x + 45$ is an increasing function when:

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