Intervals, Equalities and Functions - Part 2

Let $f(x)=ax^{3}+bx^{2}+cx+d$ . Now $f'''(x)=24$ . This derivative must come from the $ax^{3}$ term.

Now, in $f'(x)$ , the term becomes $3ax^2$ . In $f''(x)$ , it becomes $6ax$ , and becomes $6a$ in $f'''(x)$ . Thus, $6a=24\Rightarrow a=4$ .

Now, using this value for $a$ , and the third clue, the expression becomes $f(x)=4x^3+bx^2+6$

Now, $f''(x)=12x+2b=2(12x+b)$ . It is given that $\frac{-1}{6}$ is an extreme value other than the limits of the interval. Thus equating,

$f''(\frac{-1}{6})=0=2(b-2)\Rightarrow b=2$

Now, plugging into $f'(x)=12x^2+4x=4x(3x+1)$ , thus two extreme values of $x=0, \frac{-1}{3}$ . Plugging in these extreme values and the interval limits into $f(x)$ , we find the maximum when $x=2$ . At $x=2, f(x)=\boxed{46}$

Since the coefficients of $x$ and $x^0$ are 0 and 6 respectively, let $f(x) = ax^3+bx^2+6$ . Then

$\begin{aligned} f(x) & = ax^3+bx^2 + 6 \\ f'(x) & = 3ax^2 + 2bx \\ f''(x) & = 6ax+2b \\ f'''(x) & = 6a = 24 \end{aligned}$

Since $f'''(x) = 6a = 24 \implies a = 4$ . Since an extreme of $f'(x)$ occurs when $x = -\frac 16$ , this implies that $f''\left(-\frac 16\right) = 6a\left(-\frac 16\right)+2b = - 4 + 2b = 0$ $\implies b = 2$ . Therefore, $f(x) = 4x^3+2x^2+6$ and $f'(x) = 12x^2 + 4x>0$ for $x>0$ . Therefore, $f(x)$ is an increasing function for $x>0$ and it is maximum when $x=2$ , implying $\max (f(x)) = f(2) = 4(2^3) + 2(2^2) + 6 = \boxed{46}$ .