Find the intervals in which the function f(x) | vedclass

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(A) Given function is $f(x) = 10 - 6x - 2x^2$. First,find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$. To find the critic

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Strictly increasing on $(-\infty, -3/2)$ and strictly decreasing on $(-3/2, \infty)$

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(A) Given function is $f(x) = 10 - 6x - 2x^2$. First,find the derivative $f'(x)$: $f'(x) = \frac{d}{dx}(10 - 6x - 2x^2) = -6 - 4x$. To find the critical points,set $f'(x) = 0$: $-6 - 4x = 0 \Rightarrow 4x = -6 \Rightarrow x = -\frac{3}{2}$. The point $x = -\frac{3}{2}$ divides the real line into two intervals: $(-\infty, -\frac{3}{2})$ and $(-\frac{3}{2}, \infty)$. Case $1$: For $x \in (-\infty, -\frac{3}{2})$,let $x = -2$. $f'(-2) = -6 - 4(-2) = -6 + 8 = 2 > 0$. Thus,$f(x)$ is strictly increasing on $(-\infty, -\frac{3}{2})$. Case $2$: For $x \in (-\frac{3}{2}, \infty)$,let $x = 0$. $f'(0) = -6 - 4(0) = -6 Thus,$f(x)$ is strictly decreasing on $(-\frac{3}{2}, \infty)$.

Find the intervals in which the function $f(x) = 10 - 6x - 2x^2$ is strictly increasing or strictly decreasing.

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