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

Question

What is the relation between $f'(x)$ and discriminant of a quadratic equation?

Answer 1

$f(x)$ is decreasing if $f'(x)$ is negative for all $x$. So in the quadratic equation $f'(x)=0$ we need the parabola to open downward and not be positive for any $x$.

So in $f'(x)$ we need $\Delta \le 0$ and $a\lt 0$. We get $a\in (-\infty, -2]$.

Answer 2

I got the same answer as King Tut, but with a different approach:

$f'(x) = 3 \{ ax^2 - (2a+4)x + (3a+6) \}$
$f"(x) = 3 \{ 2ax - 2a - 4\} = 6 \{ax - a - 2\} = 6 \{a(x-1) - 2\}$
$f"'(x) = 6a.$

$f'(0) < 0$ if and only if $a < -2$.
$a<-2$ implies that for all $x<0,$ $f"(x) > 0.$

Thus, with $a<-2$, $f'(x)$ starts out negative at $0$ and
$f'(x)$ is monotonically decreasing as x goes from $0$ to $-\infty.$

Continuing to assume that $a<-2,$ the analysis is similar for $x>0.$
When $(x-1)=\displaystyle\frac{2}{a},$ $f"(x) = 0$ and $f"'(x) = 6a < 0.$

Therefore, $f'(x)$ has a maximum at $x=1 + \displaystyle\frac{2}{a}.$

$f'\left(1+\displaystyle\frac{2}{a}\right) = 3\left[a\left(1+\displaystyle\frac{4}{a}+\displaystyle\frac{4}{a^2}\right) - (2a+4)\left(1+\displaystyle\frac{2}{a}\right) +(3a+6)\right]$

$= 3\left[\left(a+4+\displaystyle\frac{4}{a}\right) - \left(2a+4+4+\displaystyle\frac{8}{a}\right) +(3a+6)\right]$

$= 3\left[2a+2-\displaystyle\frac{4}{a}\right] < 3[2a+2+2] < 0.$