For which interval is the given function f(x) | vedclass

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(D) Given function: $f(x) = -2x^3 - 9x^2 - 12x + 1$ Find the derivative: $f'(x) = \frac{d}{dx}(-2x^3 - 9x^2 - 12x + 1) = -6x^2 - 18x - 12$ For the fun

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

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(D) Given function: $f(x) = -2x^3 - 9x^2 - 12x + 1$ Find the derivative: $f'(x) = \frac{d}{dx}(-2x^3 - 9x^2 - 12x + 1) = -6x^2 - 18x - 12$ For the function to be decreasing,we must have $f'(x) $-6x^2 - 18x - 12 Divide by $-6$ (note that the inequality sign reverses): $x^2 + 3x + 2 > 0$ Factor the quadratic expression: $(x + 2)(x + 1) > 0$ Using the sign scheme method,the inequality $(x + 2)(x + 1) > 0$ holds when $x  -1$. Thus,the function is decreasing in the interval $(-\infty, -2) \cup (-1, \infty)$.

For which interval is the given function $f(x) = -2x^3 - 9x^2 - 12x + 1$ decreasing?

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