For what value of a is the function f(x) = (a | vedclass

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(A) Given $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$.For $f(x)$ to be a decreasing function for all $x \in R$,we must have $f'(x) \leq 0$ for all $x \in R$

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$a < -3$

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(A) Given $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$.For $f(x)$ to be a decreasing function for all $x \in R$,we must have $f'(x) \leq 0$ for all $x \in R$.$f'(x) = 3(a + 2)x^2 - 6ax + 9a$.For $f'(x) \leq 0$ for all $x \in R$,the coefficient of $x^2$ must be negative and the discriminant $D$ must be less than or equal to $0$.$1$) $a + 2 $2$) $D = (-6a)^2 - 4(3(a + 2))(9a) \leq 0$.$36a^2 - 108a(a + 2) \leq 0$.$36a^2 - 108a^2 - 216a \leq 0$.$-72a^2 - 216a \leq 0$.Divide by $-72$ (reversing the inequality sign):$a^2 + 3a \geq 0$.$a(a + 3) \geq 0$.This implies $a \in (-\infty, -3] \cup [0, \infty)$.Considering both conditions $a < -2$ and $a \in (-\infty, -3] \cup [0, \infty)$,the intersection is $a \in (-\infty, -3]$.

For what value of $a$ is the function $f(x) = (a + 2)x^3 - 3ax^2 + 9ax - 1$ a decreasing function for all $x \in R$?

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