Maxima

Find the derivative of g(x) :

$\begin{aligned} g'(x) & = -106 \cdot (3x^{4}+8x^{3}-18x^{2}+60)^{-2} \cdot (12x^{3}+24x^{2}-36x)\\ g'(x) & = \frac{-1272 \cdot x \cdot (x^{2}+2x-3)} {(3x^{4}+8x^{3}-18x^{2}+60)^{2}} \end{aligned}$

Find when g(x) = 0 (zero slope).

$\begin{aligned} 0 & = \frac{-1272 \cdot x \cdot (x^{2}+2x-3)} {(3x^{4}+8x^{3}-18x^{2}+60)^{2}} \\ 0 & = -1272 \cdot x \cdot (x^{2}+2x-3)\\ 0 & = x, 0 = x^{2}+2x-3\\ \therefore x & = -3, 0, 1 \end{aligned}$

Since the domain is x > 0 , we will need to find solutions when x=0 (critical point at the edge of the domain) and when x=1 (zero slope within the domain).

$\begin{aligned} g(0) & = 0\\ g(1) & = 2 \end{aligned}$

Then we check the right side of x = 1 , we will use x = 2 :

$\begin{aligned} g(2) & = \frac{106}{100}\\ & = 1.06 \end{aligned}$

From above, we conclude that g(0) < g(1) and 'g(2) < g(1)'. Therefore, the maximum value of domain x > 0 exists at x = 1 , where g(1) = 2 .

The answer is 2

Let f (x)= the denominator of g (x). $f'(x) = 12x (x+3)(x-1) ; x> 0$ Therefore local minima occurs at x = 1 . $f (1)= 53. g (x)_{max} = 106/53 = 2$