applications of derivatives

Question

The set of all values of '$a$' for which $f(x) = (a+2)x^3 - 3ax^2 + 9ax - 1$ decreases for all real $x$ is:

RankGuru · Accepted Answer

For $f(x)$ to decrease for all $x$, $f'(x)$ must be $\le 0$ for all $x$.
$f'(x) = 3(a+2)x^2 - 6ax + 9a$.
Condition 1: Coeff of $x^2 < 0 \implies 3(a+2) < 0 \implies a < -2$.
Condition 2: Discriminant $D \le 0$.
$D = (-6a)^2 - 4 * 3(a+2) * 9a$
$= 36a^2 - 108a(a+2)$
$= 36a^2 - 108a^2 - 216a$
$= -72a^2 - 216a$
$= -72a(a+3)$.

We need $D \le 0$:
$-72a(a+3) \le 0 \implies a(a+3) \ge 0$.
This implies $a$ is in $(-\infty, -3] \cup [0, \infty)$.

Combine with $a < -2$:
Intersection of $(-\infty, -2)$ AND $((-\infty, -3] \cup [0, \infty))$ is $(-\infty, -3]$.
Thus, $a$ belongs to $(-\infty, -3]$.

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