POSSIBLE SOLUTION(S)

Example 1

Input := 

f[x_] = 2 x^3 - 3 x^2 + x + 4;

Input := 

Plot[f[x],{x,-3,1.5}];

Input := 

Plot[f[x],{x,0,1}];

Description: This function has a value of 4 when x is 0. It decreases rapidly when x<0, becoming negative before x gets to -1. As x increases from 0, the function increases just a little bit (not more than 4.1), then decreases a little bit (not less than 3.9), all before x gets to 1. When x is 1/2 and when x is 1, the function value is 4 again, and then, for x>1, the function increases rapidly in value.

Example 2

Input := 

g[x_] = 100 - 100 Exp[-x/2];

Input := 

Plot[g[x],{x,-1,10}];

Description: This function is always increasing. Its value is 0 when x is 0, and decreases sharply as x gets more and more negative. The function has a horizontal asymptote at y=100.

This description would probably have to be refined in order for the function fitters to obtain the coefficient of x in the exponent.

Example 3

Input := 

h[x_] = 3 + 2 Sin[2x - Pi/4];

Input := 

Plot[h[x],{x,0,2Pi}];

Input := 

Plot[h[x],{x,-.01,.01}];

Input := 

h[0] //N

Output =

1.58579

Description: This function oscillates between 1 and 5. It repeats itself every Pi units. When x is 0, the value of the function is about 1.58579.

In this case, I would encourage students not to use the exact value of the function at this point (3 - Sqrt[2]), partly because data doesn't look like this, and partly because it gives away the `shape' of the function.