Question 12

If $f(x)$ were a function defined over the set of $\mathbb {R}$ minus $-\sqrt[3]{200}$ (let's pretend so), then:

$f'(x) = \displaystyle\frac{{2x({x^3} + 200) - {x^2}(3{x^2} + 200)}}{{{{({x^3} + 200)}^2}}} = \displaystyle\frac{{ - {x^4} + 400x}}{{{{({x^3} + 200)}^2}}} = \displaystyle\frac{{ - x({x^3} - 400)}}{{{{({x^3} + 200)}^2}}}$

$f'(x)=0 \Leftrightarrow x=0, x=\sqrt[3] {400} \approx 7.37$ . Hence the function $f(x)$ has two extrema: a minimum at $x = 0$ and a maximum at $x = \sqrt[3] {400}$

However, as $f(x)$ is only defined over the set of natural numbers, the maximum value of $f$ must occur at either of the two natural numbers preceding and following the value of $\sqrt[3]{400}$ , which are 7 and 8 respectively.

Plugging these values into the function, we see that:

$\begin{array}{l} f(7) = \displaystyle\frac{{49}}{{543}} \approx 0.0902 \\ f(8) = \displaystyle\frac{8}{{89}} \approx 0.0899 \\ \end{array}$

So, the maximum value of $f(x)$ is $f(7) = \boxed {\displaystyle\frac{{49}}{{543}}}$