f(x) = 10 - 6x - 2x^2 is strictly increasing | vedclass

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(A) To find the interval where the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing,we first find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(

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$(-\infty, -\frac{3}{2})$

Vedclass · Answer

(A) To find the interval where the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing,we first find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$.For the function to be strictly increasing,we must have $f'(x) > 0$.$-6 - 4x > 0$$-4x > 6$Dividing by $-4$ reverses the inequality sign:$x $x Thus,the function is strictly increasing in the interval $(-\infty, -\frac{3}{2})$.Therefore,the correct option is $A$.

$f(x) = 10 - 6x - 2x^2$ is strictly increasing in the . . . . . . interval.

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