The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a positive point of maximum lies in the interval $(a_1,a_2)\cup(a_3,a_4)$ 
The value of $a$ for which $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ have a
  positive point of maximum lies in the interval
  $(a_1,a_2)\cup(a_3,a_4)$.find the value of $a_2+11a_3+70a_4$

I differentiated the equation $f(x)=x^3+3(a-7)x^2+3(a^2-9)x-1$ and put it equal to zero, to get $f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$. Now what to do to get the desired interval. What is the significance of positive point of maximum in this question.
Please help. Thanks in advance.
 A: HINT:
The solution to the equation
$$f'(x)=3x^2+6(a-7)x+3(a^2-9)=0$$
is 
$$\frac{-6(a-7)\pm \sqrt{(6a-42)^2-4\cdot 3 \cdot 3(a^2-9)}}{2\cdot 3}$$
If there exists a point of maximum, then you have
$$(6a-42)^2-4\cdot 3 \cdot 3(a^2-9)\geq 0$$
Moreover, since this point is positive, you have
$$\sqrt{(6a-42)^2-4\cdot 3 \cdot 3(a^2-9)}\geq 6(a-7)$$
Solving those equations should give you the desired intervals.
A: if you say 
$$f''(x)=6x+6(a-7)
$$
then all of the possible values where this is less than zero gives you the range of the possible maximum values. you can find all of the possible values for x in terms of 'a' using the quadratic formula on the first derivative $f'(x)$. 
Once you have these values for 'x' you can sub them all into the second derivative $f''(x)$ and find the range of 'a' values which will make this less than zero. This range should be comparable to the interval you have stated with a maximum and minimum x and y value.
I assume these values are the values fo $a_1,a_2,a_3,a_4$ and you then solve from there
A: The zeroes of $f'$ are $-a+7\pm\sqrt{-14a+58}$.  Now determine $a$ so that the zeroes are zero, you'll find $a=\pm3$.  From here you'll be able to find the desired intervals.
