The function y = 6 - 9x - x^2 is strictly inc | vedclass

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(A) To determine the interval where the function $y = f(x) = 6 - 9x - x^2$ is strictly increasing,we find its derivative: $f'(x) = \frac{d}{dx}(6 - 9x

Vedclass · Accepted Answer

$(-\infty, -4.5)$

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(A) To determine the interval where the function $y = f(x) = 6 - 9x - x^2$ is strictly increasing,we find its derivative: $f'(x) = \frac{d}{dx}(6 - 9x - x^2) = -9 - 2x$. For the function to be strictly increasing,we must have $f'(x) > 0$. So,$-9 - 2x > 0$. $-2x > 9$. Dividing by $-2$ reverses the inequality: $x Thus,the function is strictly increasing on the interval $(-\infty, -4.5)$.

The function $y = 6 - 9x - x^2$ is strictly increasing on the interval . . . . . . .

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