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

Hint: In this question it is given that we have to find the value of a for which the function $$f\left( x\right) =ax^{3}-3\left( a+2\right) x^{2}+9\left( a+2\right) x-1$$ is decreasing for all $$x\in \mathbf{R}$$.
So to find the solution we need to know that if a function is decreasing in any domain then the first derivative of that function in that domain is always greater than zero, i.e, $$f^{\prime }\left( x\right) < 0$$.

Complete step-by-step answer:
Given function,
$$f\left( x\right) =ax^{3}-3\left( a+2\right) x^{2}+9\left( a+2\right) x-1$$..........(1)
Now differentiating both side w.r.t ‘x’ we get,
$$\dfrac{d}{dx} f\left( x\right) =\dfrac{d}{dx} \left\{ ax^{3}-3\left( a+2\right) x^{2}+9\left( a+2\right) x-1\right\} $$
$$\Rightarrow f^{\prime }\left( x\right) =a\dfrac{d}{dx} \left( x^{3}\right) -3\left( a+2\right) \dfrac{d}{dx} \left( x^{2}\right) +9\left( a+2\right) \dfrac{d}{dx} \left( x\right) -\dfrac{d}{dx} \left( 1\right) $$
[$$\because \dfrac{d}{dx} f\left( x\right) =f^{\prime }\left( x\right) $$]
Now as we know that $$\dfrac{d}{dx} \left( x^{n}\right) =nx^{n-1}$$,
So we can write,
$$f^{\prime }\left( x\right) =a\times 3x^{2}-3\left( a+2\right) \times 2x+9\left( a+2\right) \times 1-0$$
$$=3ax^{2}-6\left( a+2\right) x+9\left( a+2\right) $$........(2)
Here it is given $$x\in \mathbf{R}$$, i.e, $$-\infty < x <\infty$$
Now since the function is decreasing therefore we can write,
$$f^{\prime }\left( x\right) <0$$
$$\Rightarrow 3ax^{2}-6\left( a+2\right) x+9\left( a+2\right)
<0$$.......(3)
So for this we have to know that if any quadratic equation is less than zero, then the coefficient of x is less than zero also discriminant (D) is also less than zero,
i.e, if $$px^{2}+qx+r < 0$$ then p < 0 and D < 0.
Where D = $$q^{{}2}-4pr$$.
Now if we compare the equation (3) with $$px^{2}+qx+r<0$$, then we can write,
p = 3a, q = -6(a+2), r = 9(a+2)
Therefore by the above formula,
 p < 0
$$\Rightarrow 3a < 0$$
$$\Rightarrow a < 0$$
Therefore, $$a\in \left( -\infty ,0\right) $$......(4)
And,
$$q^{2}-4pr<\ 0$$
$$\Rightarrow \left\{ -6\left( a+2\right) \right\}^{2} -4\times 3a\times 9\left( a+2\right) <\ 0$$
$$\Rightarrow 36\left( a+2\right)^{2} -108a\left( a+2\right) <\ 0$$
$$\Rightarrow \left( a+2\right) \{ 36\left( a+2\right) -108a\} <\ 0$$
$$\Rightarrow \left( a+2\right) \{ 36a+72-108a\} <\ 0$$
$$\Rightarrow \left( a+2\right) (72-72a)<\ 0$$
$$\Rightarrow \left( a+2\right) \times 72\left( 1-a\right) <\ 0$$
$$\Rightarrow 72\left( a+2\right) \left( 1-a\right) <\ 0$$
$$\Rightarrow -72\left( a+2\right) \left( a-1\right) <\ 0$$
$$\Rightarrow -\left( a+2\right) \left( a-1\right) <\ 0$$
$$\Rightarrow \left( a+2\right) \left( a-1\right) >\ 0$$ [if -a < b, then a > -b]
$$\text{Either} \ \left( a+2\right) >0\ \text{and} \ \left( a-1\right) >0$$
$$\therefore a>-2\ \text{and} \ a>1$$
Which gives, $$a\in \left( 1,\infty \right) $$
$$\text{or} \ \left( a+2\right) <0\ \text{and} \ \left( a-1\right) <0$$
$$\therefore a<-2\ \text{and} \ a<1$$
Which implies, $$a\in \left( -\infty ,-2\right) $$
Now since in between the conditions or is given so we have to take union, i,e, $$a\in \left( -\infty ,-2\right) \cup \left( 1,\infty \right) $$..........(5)
Now from we have to take intersection of a from the conditions (4) and (5),
Therefore we get,
$$a\in \left( -\infty ,0\right) \cap \left\{ \left( -\infty ,-2\right) \cup \left( 1,\infty \right) \right\} $$
Which implies, $$a\in \left( -\infty ,-2\right) $$,
Which gives the all possible values of a for which the function is decreasing.


Note: While solving this type of question you need to know that if you have given ab > 0, then from here two cases arise and the cases are either, a > 0 & b > 0 or a < 0 & b < 0., i,e, multiplication of two factors is positive if and only if they both positive or they both negative.