At the beginning the function is  F: x → x^3−2, that I could study with this program, for polynomial functions.

```function F(x) {
with(Math) {
/// you can change F [ now it is  x - > x^3-2, ie pow(x,3)-2 ]

return pow(x,3)-2

}}
```

To consider another function, we study F: x → x³·cos(x)/( (cos(x)+x)·(sin(x)−x) ) in the interval [-0.5, 5].

```function F(x) {
with(Math) {

y = pow(x,3)*cos(x) / (  (cos(x)+x)*( sin(x)-x ) )
return y

}}```

(we can draw the graphs with JavaScript; for example, see the script x3cossin  - see the code;  se also here).

In x=0, where sin(x)-x = 0, it is not defined. It is continuous function in [-1/2,0) ∪ (0,5]. How can we fill the hole in order to obtain a continuous function in [-1/2,5]? We calculate lim x → 0 F(x). We can use this script (to tabulate the function), with the new F:

```function F(x) { with(Math) {
y = pow(x,3)*cos(x) / (  (cos(x)+x)*( sin(x)-x ) )
return y
}}```

We could thicken the tab, but it is "evident" that the limit is 6.

Let's look for the "maximum" of F. With this program (with the new F) I find:

F has the maximum value 4.65229469036 at 3.015673948.